Chapter 9. Answers of Straight Lines

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Since the intercept and angle with $x$ -axis is given, therefore, we can represent this line using slope intercept form i.e. $y = m x + c$ .

$m = \tan 45^{\circ} = 1$ and $c = 5$ . Therefore, the required equation is $y = x + 5$ .

Let the interecepts be $a, - a$ for the two axes. So we can represent the line as $x / a - y / a = 1$ . Since it passes through $(2, - 3) \Rightarrow \frac{2}{a} - \frac{- 3}{a} = 1 \Rightarrow a = 5$ .

Thus, the required equation is $x - y = 5$ .

Let the required straight line be $\frac{x}{a} + \frac{y}{b} = 1$ , which will meet the axes at $(a, 0)$ and $(0, b)$ .

The coordinate of the point dividing the line joining these points in the ratio $1 : 2$ is $\frac{2. a + 1.0}{2 + 1}$ and $\frac{2.0 + 1. b}{2 + 1}$ i.e. $\frac{2 a}{3}$ and $\frac{b}{3}$ .

Thus, $\frac{2 a}{3} = - 5 \Rightarrow a = - \frac{15}{2}$ and $4 = \frac{b}{3} \Rightarrow b = 12$ .

Thus, the required equationis $\frac{x}{- \frac{15}{2}} + \frac{y}{12} = 1 \Rightarrow 5 y - 8 x = 60$ .

Comparing the equation with $a x + b y + c = 0$ we have $a = 1, b = \sqrt{3}, c = 1$

$\sqrt{a^{2} + b^{2}} = \sqrt{1 + 3} = 2$

Dividing the given equation by $2$ gives us

$\frac{x}{2} + y . \frac{\sqrt{3}}{2} + \frac{7}{2} = 0 \Rightarrow x \cos 240^{\circ} + y \sin 240^{\circ} = \frac{7}{2}$ .

The equation of straight in two-point form is given by $y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$

So the equation becomes $y - 3 = \frac{- 2 - 3}{4 + 1} (x + 1) \Rightarrow x + y = 2$ .

Since the intercept and slope is given we can represent it as $y = m x + c$ . Given that $c = 1$ and $m = \tan 45^{\circ} = 1$ .

Therefore, the required equation is $y = x + 1$ .

Since the intercept and slope is given we can represent it as $y = m x + c$ . Given that $c = - 5$ . Since the line is equally inclined to the axes so angle of inclination is $θ = 45^{\circ}$ .

Thus, $m = \tan θ = 1$ , and thus, the required equation is $y = x - 5$ .

Since the intercept and slope is given we can represent it as $y = m x + c$ . Given that $c = - 2$ . Since the line is inclined at $30^{\circ}$ to $O X$ so $m = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ .

Thus, required equation is $y = \frac{x}{\sqrt{3}} - 2 \Rightarrow \sqrt{3} y = x - 2 \sqrt{3}$ .

Since the intercept and slope is given we can represent it as $y = m x + c$ . Given that $c = - 3$ . Since the line is inclined at an angle $\tan^{- 1} \frac{3}{5}$ , therefore $m = \tan \tan^{- 1} \frac{3}{5} = \frac{3}{5}$ .

Thus, required equation is $y = \frac{3}{5} x - 3 \Rightarrow 5 y = 3 x - 9$ .

Since intercepts are given so we can use the intercept form i.e. $\frac{x}{a} + \frac{y}{b} = 1$ . Given that $a = 2$ ande $b = 3$ .

Therefore, the equation of the line is $\frac{x}{2} + \frac{y}{3} = 1 \Rightarrow 3 x + 2 y = 6$ .

Since intercepts are given so we can use the intercept form i.e. $\frac{x}{a} + \frac{y}{b} = 1$ . Given that $a = - 5$ and $b = 6$ .

Therefore, the equation of the line is $\frac{x}{- 5} + \frac{y}{6} = 1 \Rightarrow 5 y = 6 x + 30$ .

Let the intercept be $a$ on both the axes. Then the equation of the line would be $\frac{x}{a} + \frac{y}{a} = 1 \Rightarrow x + y = a$ . Since the line passes through $(5, 6)$ , therefore, $a = 5 + 6 = 11$ . And thus, the equation of the straight line is $x + y = 11$ .

In the second case let the intercepts be $a, - a$ . Then the equation would be $\frac{x}{a} - \frac{y}{a} = 1 \Rightarrow x - y = a$ . Since the line passes through $(5, 6)$ , therefore, $a = 5 - 6 = - 1$ . And thus, equation of the straight line would be $x - y + 1 = 0$ .

First let the intercepts be $a, a$ , then the equation of the line would be $\frac{x}{a} + \frac{y}{a} = 1 \Rightarrow x + y = a$ . Since the line passes theough $(1, - 2)$ , therefore, $a = 1 - 2 = - 1$ . And thus, the equation fo the straight line is $x + y = - 1$ or $x + y + 1 = 0$ .

Now let the intercepts be $a, - a$ , then $a = x - y = 1 + 2 = 3$ . So, the equation would be $x - y = 3$ .

Let $a, b$ are the intercepts with $1$ -axis and $y$ -axis respectively. Since $(x^{'}, y^{'})$ bisects it therefore $x^{'} = \frac{a}{2} => a = 2 x^{'}$ , and similarly, $b = 2 y^{'}$ . The equation of line would be $x / a + y / b = 1$ .

And thus, the required equation of the line in question is $x y^{'} + x^{'} y = 2 x^{'} y^{'}$ .

Let $a, b$ are the intercepts with $x$ -axis and $y$ -axis respectively. Since $(- 4, 3)$ divides the intercept in the ratio $5 : 3$ , therefore, $- 4 = \frac{5.0 + 3. a}{5 + 3} \Rightarrow a = - \frac{32}{3}$ and $3 = \frac{5. b + 3.0}{5 = 3} \Rightarrow b = \frac{24}{5}$ .

And thus equation of line is $\frac{x}{a} + \frac{y}{b} = 1 \Rightarrow - \frac{3 x}{32} + \frac{5 y}{24} = 1 \Rightarrow 20 y - 9 x = 96$ .

We will make use of two point form. The equation of the line is given by

$y - 0 = \frac{- 2 - 0}{2 - 0} (x - 0) \Rightarrow x + y = 0$ .

We will make use of two point form. The equation of the line is given by

$y - 4 = \frac{6 - 4}{5 - 3} (x - 3) \Rightarrow y = x + 1$ .

We will make use of two point form. The equation of the line is given by

$y - 3 = \frac{- 7 - 3}{6 + 1} (x + 1) \Rightarrow 7 y - 21 = - 10 x - 10 \Rightarrow 10 x + 7 y = 11$ .

This problem can be solved with intercept form. Intercept on $x$ -axis is $b$ and on $y$ -axis is $- a$ . Thus, equation of the line is $\frac{x}{b} - \frac{y}{a} = 1 \Rightarrow a x - b y = a b$ .

We will make use of two point form. The equation of the line is given by

$y - b = \frac{a - b - b}{a + b - a} (x - a) \Rightarrow b y - b^{2} = (a - 2 b) x - a^{2} + 2 a b \Rightarrow (a - 2 b) x - b y + b^{2} + 2 a b - a^{2} = 0$ .

The equation of the given line is given by $y - 2 a t_{1} = \frac{2 a t_{2} - 2 a t_{1}}{a t_{2}^{2} - a t_{1}^{2}} (x - a t_{1}^{2}) = \frac{2}{t_{2} + t_{1}} (x - a t_{1}^{2})$

$\Rightarrow y (t_{1} + t_{2}) - 2 x = 2 a t_{1} t_{2}$ .

The equation of the given line is given by $y - a t_{1} = \frac{\frac{a}{t_{2}} - \frac{a}{t_{1}}}{a t_{2} - a t_{1}} (x - a t_{1}) = - \frac{1}{t_{1} t_{2}} (x - a t_{1})$

$\Rightarrow t_{1} t_{2} y + x = a (t_{1} + t_{2})$ .

The equation of the line is given by $y - a \sin ϕ_{1} = \frac{a \sin ϕ_{2} - a \sin ϕ_{1}}{a \cos ϕ_{2} - a \cos ϕ_{1}} (x - a \cos ϕ_{1})$

$\Rightarrow y - a \sin ϕ_{1} = \frac{2 \cos \frac{ϕ_{1} + ϕ_{2}}{2} \sin \frac{ϕ_{2} - ϕ_{1}}{2}}{2 \sin \frac{ϕ_{1} + ϕ_{2}}{2} \sin \frac{ϕ_{1} - ϕ_{2}}{2}} (x - a \cos ϕ_{1})$

$=> x \cos \frac{ϕ_{1} + ϕ_{2}}{2} + y \sin \frac{ϕ_{1} + ϕ_{2}}{2} = a \cos \frac{ϕ_{1} - ϕ_{2}}{2}$ .

The equation of the line is given by $y - b \sin ϕ_{1} = \frac{b \sin ϕ_{2} - b \sin ϕ_{1}}{a \cos ϕ_{2} - a \cos ϕ_{1}} (x - a \cos ϕ_{1})$

$=> y - b \sin ϕ_{1} = \frac{2 b c o s \frac{ϕ_{1} + ϕ_{2}}{2} \sin \frac{ϕ_{2} - ϕ_{1}}{2}}{2 a \sin \frac{ϕ_{1} + ϕ_{2}}{2} \sin \frac{ϕ_{1} - ϕ_{2}}{2}} (x - a \cos ϕ_{1})$

$\Rightarrow \frac{x}{a} \cos \frac{ϕ_{1} + ϕ_{2}}{2} + \frac{y}{b} \sin \frac{ϕ_{1} + ϕ_{2}}{2} = \cos (ϕ_{1} - ϕ_{2}) / 2$ .

The equation of the line is given by $y - b \tan ϕ_{1} = \frac{b \tan ϕ_{2} - b \tan ϕ_{1}}{a \sec ϕ_{2} - a \sec ϕ_{1}} (x - a \sec ϕ_{1})$

Now $\frac{b \tan ϕ_{2} - b \tan ϕ_{1}}{a \sec ϕ_{2} - a \sec ϕ_{1}} = \frac{b}{a} . \frac{\sin ϕ_{2} \cos ϕ_{1} - \sin ϕ_{1} \cos ϕ_{2}}{\cos ϕ_{1} - \cos ϕ_{2}}$

$= \frac{\sin (ϕ_{2} - ϕ_{1})}{\cos ϕ_{1} - \cos ϕ_{2}} = \frac{2 \sin \frac{ϕ_{2} - ϕ_{1}}{2} \cos \frac{ϕ_{2} - ϕ_{1}}{2}}{2 \sin \frac{ϕ_{1} + ϕ_{2}}{2} \sin \frac{ϕ_{2} - ϕ_{1}}{2}}$

Simplifying gives us the equation of the line as

$b x \cos \frac{ϕ_{1} - ϕ_{2}}{2} - a y \sin \frac{ϕ_{1} + ϕ_{2}}{2} = a b \cos \frac{ϕ_{1} + ϕ_{2}}{2}$ .

Let the vertices of the triangle be $A (1, 4), B (2, - 3)$ and $C (- 1, - 2)$ .

We find the equations of the sides $A B, B C$ and $C A$ .

Slope of $A B : m_{A B} = \frac{- 3 - 4}{2 - 1} = - 7$

Equation of $A B : y - 4 = - 7 (x - 1)$ so $y = - 7 x + 11$

Slope of $B C : m_{B C} = \frac{- 2 - (- 3)}{- 1 - 2} = - \frac{1}{3}$

Equation of $B C : y + 3 = - \frac{1}{3} (x - 2)$ so $y = - \frac{1}{3} x - \frac{7}{3} \Rightarrow 3 y + x + 7 = 0$

Slope of $C A : m_{C A} = \frac{4 - (- 2)}{1 - (- 1)} = 3$

Equation of $C A : y - 4 = 3 (x - 1)$ so $y = 3 x + 1$

Hence, the equations of the sides are: $A B : y + 7 x = 11, B C : 3 y + x + 7 = 0$ and $C A : y = 3 x + 1$

Let the vertices of the triangle be $A (0, 1), B (2, 0)$ and $C (- 1, - 2)$ .

We find the equations of the sides $A B, B C$ and $C A$ .

Slope of $A B : m_{A B} = \frac{0 - 1}{2 - 0} = - \frac{1}{2}$

Equation of $A B : y - 1 = - \frac{1}{2} (x - 0)$ so $y = - \frac{1}{2} x + 1$

Slope of $B C : m_{B C} = \frac{- 2 - 0}{- 1 - 2} = \frac{2}{3}$

Equation of $B C : y - 0 = \frac{2}{3} (x - 2)$ so $y = \frac{2}{3} x - 4 / 3$

Slope of $C A : m_{C A} = \frac{1 - (- 2)}{0 - (- 1)} = 3$

Equation of $C A : y - 1 = 3 (x - 0)$ so $y = 3 x + 1$

Hence, the equations of the sides are: $A B : 2 y + x = 2, B C : y = 2 / 3 x - 4 / 3$ , and $C A : y = 3 x + 1$ .

Intersection of $x = a, y = b$ will give the point $(a, b)$ and opposite to it will be intersection of the lines $x = a^{'}$ and $y = b^{'}$ i.e. $(a^{'}, b^{'})$ . Equation of this diagonal would be $y - b = \frac{b^{'} - b}{a^{'} - a} (x - a) \Rightarrow (b^{'} - b) x - (a^{'} - a) y + (a^{'} - a) b + (b - b^{'}) a = (b^{'} - b) x + (a - a^{'}) y + a^{'} b - a b^{'}$ .

Intersection of $x = a, y = b^{'}$ will give the point $(a, b^{'})$ and opposite to it will be intersection of the lines $x = a^{'}$ and $y = b$ i.e. $(a^{'}, b)$ . Equation of this diagonal would be $y - b = \frac{b - b^{'}}{a^{'} - a} (x - a^{'})$ , simplification of which is left to you.

Figure 9.1.

Point which bisects the distance between $(a, b)$ and $(a^{'}, b^{'})$ is given by $(\frac{a + a^{'}}{2}, \frac{b + b^{'}}{2})$ , and point which bisects the distance between $(- a, b)$ and $(a^{'}, - b^{'})$ is given by $(\frac{a^{'} - a}{2}, \frac{b - b^{'}}{2})$ .

The equation of the line passing through these points obtained is given by $y - \frac{b + b^{'}}{2} = (\frac{\frac{b - b^{'}}{2} - \frac{b + b^{'}}{2}}{\frac{a^{'} - a}{2} - \frac{a + a^{'}}{2}}) (x - \frac{a + a^{'}}{2})$

$\Rightarrow 2 a y - 2 b^{'} x = a b - a^{'} b^{'}$ .

Intercepts of the line $3 x + y = 12$ are $(4, 0)$ and $(0, 12)$ . The points which trisect these lines are $(\frac{2.4 + 0.1}{3}, \frac{0.2 + 12.1}{3})$ and $(\frac{1.4 + 0.2}{3}, \frac{0.1 + 12.2}{3})$ i.e. $(\frac{8}{3}, 4)$ and $(\frac{4}{3}, 8)$ .

Line passing through origin and $(\frac{8}{3}, 4)$ is $y = \frac{4}{\frac{8}{3}} x \Rightarrow 3 x = 2 y$ , and line passing through origin and $(\frac{4}{3}, 8)$ is $y = \frac{8}{\frac{4}{3}} x \Rightarrow y = 6 x$ .

Slope of the line $= m = \tan 15^{\circ} = \tan (45^{\circ} - 30^{\circ}) = (1 - \frac{1}{\sqrt{3}}) (1 + \frac{1}{\sqrt{3}}) = 2 - \sqrt{3}$

Intercept on $y$ -axis is $c = - 4$ . Therefore, the equation of the line is $y = (2 - \sqrt{3}) x - 4$ .

From the diagram it is clear that angle made with positive direction of $x$ -axis is $60^{\circ}$ . Thus, slope of the line is $m = \tan 60^{\circ} = \sqrt{3}$ , and the intercept with $y$ -axis is $- 4 \sqrt{3}$ .

Thus, equation of the line is $\sqrt{3} x - y - 4 \sqrt{3} = 0$ .

Figure 9.2.

Given that $\cos θ = - \frac{1}{3} \Rightarrow \tan θ = - \sqrt{8} = m$ , which is slope of the line. Thus, the equation of the line is given by

$y - 2 = - \sqrt{8} (x - 1) \Rightarrow 2 \sqrt{2} x + y - 2 \sqrt{2} - 2 = 0$ .

Equation of the line is given by $y - 3 = \frac{3 + 2}{- 1 - 4} (x + 1) \Rightarrow x + y = 2$ .

Given $A = (2, 0)$ , and $A B$ is the initial position, and $A C$ is the final position after rotation.

Given $∠ B A X = 30^{c} i r c$ , and $∠ B A C = 15^{\circ} ∴ ∠ C A X = 15^{\circ}$

Slope of line $A C = \tan 15^{\circ} = 2 - \sqrt{3}$

Therefore, equation of line $A C$ is $y - 0 = (2 - \sqrt{3}) (x - 2) \Rightarrow (2 - \sqrt{3}) x - y - 4 + 2 \sqrt{3} = 0$ .

Figure 9.3.

Let $A D$ be the internal bisector of the $∠ B A C$ which meets the side $B C$ at $D$ .

Now $A B = \sqrt{(5 - 2)^{2} + (2 - 3)^{2}} = \sqrt{10}$ , and $A C = \sqrt{(5 - 6)^{2} + (2 - 5)^{2}} = \sqrt{10}$

Since $A D$ is the internal bisector, therefore, $\frac{B D}{D C} = \frac{A B}{A C} = \frac{\sqrt{10}}{\sqrt{10}} = 1$

$∴ D = (\frac{2 + 6}{2}, \frac{3 + 5}{2}) = (4, 4)$

Now equation of $A D$ is $y - 2 = \frac{2 - 4}{5 - 4} (x - 5) \Rightarrow 2 x + y = 12$ .

Let $A B C D$ be a rectangle such that $A = (1, 2)$ and $C = (5, 5)$ . Clearly, vertices $B$ and $D$ lie on the line $x = 3$ . Let them be $B (3, y_{1})$ and $D (3, y_{2})$ .

Since $A C$ and $B D$ bisect each other, therefore, their middle-points will be same.

Thus, $\frac{y_{1} + y_{2}}{2} = \frac{2 + 5}{2} \Rightarrow y_{1} + y_{2} = 7$ .

Also, $B D^{2} = A C^{2} \Rightarrow (y_{1} - y_{2})^{2} = (1 - 5)^{2} + (2 - 5)^{2} = 25 \Rightarrow y_{1} - y_{2} = \pm 5$

$\Rightarrow y_{1} = 6, y_{2} = 1$ or $y_{1} = 1, y_{2} = 6$ . So the other vertices are $(3, 1)$ and $(3, 6)$ . Let $B$ represent $(3, 1)$ and $D$ represent $(3, 6)$ .

Equation of side $A B$ is $y - 2 = \frac{2 - 1}{1 - 3} (x - 1) \Rightarrow x + 2 y = 5$ .

Equation of side $B C$ is $y - 1 = \frac{1 - 5}{3 - 5} (x - 3) \Rightarrow 2 x - y = 5$ .

Equation of side $C D$ is $y - 5 = \frac{5 - 6}{5 - 3} (x - 5) \Rightarrow x + 2 y = 15$ .

Equation of side $A D$ is $y - 2 = \frac{2 - 6}{1 - 3} (x - 1) \Rightarrow 2 x = y$ .

Equation of $O T :$ Slope of $O T = \tan 45^{\circ} = 1$ and it passes through $O (0, 0)$ .

Thus, equation is $y - 0 = 1. (x - 0) \Rightarrow y = x$ .

Equation of $O S :$ Slope of $O S = \tan 135^{\circ} = - 1$ and it passes through $O (0, 0)$ .

Thus, equation is $y - 0 = - 1 (x - 0) \Rightarrow x + y = 0$ .

Equation of $S P :$ Given $O T = 2 \sqrt{2} ∴ O P = O T \sec 45^{\circ} = 4 ∴ P = (0, 4)$ .

Also, slope of the line $S P$ is $\tan 45^{\circ} = 1$ .

Thus, equation is $y - 4 = 1 (x - 0) \Rightarrow y = x + 4$ .

Equation of $Q R :$ Given $O Q = O T \sec 45^{\circ} = 4 ∴ Q = (4, 0)$ .

Slope of line $Q R = \tan 75^{\circ} = \tan (45^{\circ} + 30^{\circ}) = (\frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}) = 2 + \sqrt{3}$ .

Thus, equation is $y - 0 = (2 + \sqrt{3}) (x - 4) \Rightarrow (2 + \sqrt{3}) x - y - 8 - 4 \sqrt{3} = 0$ .

Equation of $P R : P = (4, 0)$ . Slope of line $P R = \tan 15^{\circ} = \tan (45^{\circ} - 30^{\circ}) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = 2 - \sqrt{3}$ .

Thus, equation is $y - 4 = (2 - \sqrt{3}) (x - 0) \Rightarrow (2 - \sqrt{3}) x - y + 4 = 0$ .

Equation of $P Q : P = (0, 4)$ and $Q = (4, 0)$ .

Thus, equation is $y - 4 = \frac{4 - - 0}{0 - 4} (x - 0) \Rightarrow x + y = 4$ .

Let $A D, B E$ and $C F$ meet at $O$ . We take $O$ as origin. Let the coordinates of points $A, B$ and $C$ be $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ respectively.

Let $D$ divide $B C$ in the ratio $k : 1$ i.e. $\frac{B D}{D C} = k$ then $D = (\frac{k x_{3} + x_{2}}{k + 1}, \frac{k y_{3} + y_{2}}{k + 1})$

Also, equation of line $A D$ is $y - 0 = \frac{y_{1} - 0}{x_{1} - 0} (y_{1} - 0) / (x_{1} - 0) (x - 0) \Rightarrow y = \frac{y_{1}}{x_{1}} x$

Since $D$ lies on $A D$ , therefore, $\frac{k y_{3} + y_{2}}{k + 1} = \frac{y_{1}}{x_{1}} (\frac{k x_{3} + x_{2}}{k + 1})$

$k = \frac{B D}{D C} = \frac{x_{2} y_{1} - x_{1} y_{2}}{x_{1} y_{3} - x_{3} y_{1}}$

Similarly, $\frac{C E}{E A} = \frac{x_{3} y_{2} - x_{2} y_{3}}{x_{2} y_{1} - x_{1} y_{2}}$ , and $\frac{A F}{F B} = \frac{x_{1} y_{3} - x_{3} y_{1}}{x_{3} y_{2} - x_{2} y_{3}}$

Thus, $\frac{B D}{D C} . \frac{C E}{E A} . \frac{A F}{F B} = 1$ .

Let $P Q R S$ be the square inscribed in the $△ A B C$ . Let $P = (a, 0)$ and length of each side of the square be $k$ then $S = (a + k, 0), Q = (a, k), R = (a + k, k)$ .

Equation of the line $A B$ is $y - 0 = \frac{1 - 0}{2 - 0} (x - 0) \Rightarrow x = 2 y$ .

Equation of the line $B C$ is $y - 0 = \frac{0 - 1}{3 - 2} (x - 3) \Rightarrow x + y = 3$ .

Since $Q (a, k)$ lies on $A B$ , therefore, $a = 2 k$ .

Again $R (a + k, k)$ lies on $B C$ , therefore, $a + 2 k = 3 \Rightarrow k = \frac{3}{4}, a = \frac{3}{2}$ .

Hence, $P = (\frac{3}{2}, 0), Q = (\frac{3}{2}, \frac{3}{4}), R = (\frac{9}{4}, \frac{3}{4})$ and $S = (\frac{9}{4}, 0)$ .

Figure 9.4.

Equation of the given line is $\sqrt{3} y - 3 x = 3 \Rightarrow y = \sqrt{3} x + \sqrt{3}$ , which is of the form $y = m x + c$ .

Slope of the line is $\sqrt{3} = \tan 60^{\circ}$ . Thus, the given line makes an angle of $60^{\circ}$ with the $x$ -axis.

Since slope and intercept are given, therefore, slope-intercept form can be used. Given that $m = 3, c = 7$ , therefore, equation of the straight line is $y = 3 x + 7$ .

Since slope and intercept are given, therefore, slope-intercept form can be used. Given that

$m = \tan 75^{\circ} = \tan (45^{\circ} + 30^{\circ}) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = 2 + \sqrt{3}$ , and $c = 3$ .

Therefore, the equation of the line is $y = (2 - \sqrt{3}) x + 3$ .

Since slope and intercept are given, therefore, slope-intercept form can be used. Given that

$m = \tan \sin^{- 1} \frac{12}{13} = \frac{5}{13}$ and $c = - 5$

Therefore, the equation of the line is $y = \frac{5}{13} x - 5 \Rightarrow 5 x - 13 y = 65$ .

Since the line is parallel to $x$ -axis, therefore, it will make an angle of $0^{\circ}$ with $x$ -axis i.e. $m = \tan 0^{\circ} = 0$ . Also, since its distance from $x$ -axis is $5$ units, therefore, the intercept on $y$ -axis is $5$ , which makes intercept $c = 5$ .

Thus, equation of the line would be $y = 0. x + 5 \Rightarrow y = 5$ .

Since it is not given that intercept is from positive or negative direction of $y$ -axis, therefore, the other line would be $y + 5 = 0$ .

Since the line is parallel to $y$ -axis therefore the equation would be $x = k$ , where $k$ is the intercept on $x$ -axis, which is given as $- 4$ . Therefore, the equation of the line is $x = - 4$ .

Lines parallel and perpendicular to $x$ -axis are given by $x = k$ and $y = p$ , where $k$ and $p$ are distance of the line from the $y$ -axis and $x$ -axis respectively.

Since these lines pass through $(5, 3)$ , therefore, $x = 5$ and $y = 3$ are the desired equations of the straight lines.

Since the line makes an angle of $135^{\circ}$ with positive direction of the $y$ -axis, therefore, it makes an angle of $135^{\circ}$ with positive direction of the $x$ -axis. Thus, slope of the line is $m = \tan 135^{\circ} = - 1$ .

Also give that it cuts an intercept of $2$ from positive direction of the $x$ -axis, which means that it passes through $(2, 0)$ .

Thus, equation of the straight line would be $y - 0 = - 1. (x - 2) \Rightarrow x + y = 2$ .

Since the slope is $2$ and the line cuts an intercept of $4$ on $x$ -axis i.e. it passes through $(4, 0)$ the equation of the line would be

$y = 0 = 2 (x - 4) \Rightarrow 2 x - y = 8$ .

Since the line makes an angle of $60^{\circ}$ with the positive direction of the $y$ -axis, therefore, it would make an angle of $30^{\circ}$ with the positive direction of $x$ -axis. Therefore, the slope of the line is $m = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ .

Also given that the line passes through $(3, - 2)$ , thus the equation of the line would be

$y + 2 = \frac{1}{\sqrt{3}} (x - 3) => x - \sqrt{3} y = 3 + 2 \sqrt{3}$ .

Slope is given by $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{2 - 4}{1 - 3} = 1$ .

The equation of the line would be $y - 4 = 1. (x - 3) \Rightarrow x - y + 1 = 0$ .

The equation of the line is given by $y - b = \frac{b + r \sin θ - b}{a + r \cos θ - a} (x - a)$

$\Rightarrow x \tan θ - y = a \tan θ - b$ .

The equation of the straight line is given by $y + 3 = \frac{2 + 3}{- 4 - 1} (x - 1) \Rightarrow x + y + 2 = 0$ .

Equation of the straight line passing through $(1, 4)$ and $(3, - 2)$ is given by

$y - 4 = (- 2 - 4) / (3 - 1) (x - 1) => 3 x + y = 7$ .

Now we put $(- 3, 16)$ in this equation which gives us $3. - 3 + 16 = 7$ , which is true. Thus, the point $(- 3, 16)$ also lies on the same line making the points collinear. We could have found the equation between $(3, - 2)$ and $(- 3, 16)$ which would also give the same equation.

Another way would be finding the area of the triangle whose vertices are the given three points and we will find that area of the triangle is zero; making the points collinear.

Line passing through $(a, b)$ and $(a_{1}, b_{1})$ is given by

$y - b = \frac{b_{1} - b}{a_{1} - a} (x - a) \Rightarrow (b_{1} - b) x - a (b_{1} - b) = (a_{1} - a) y - b (a_{1} - a)$

$=> (b_{1} - b) x - a b_{1} = (a_{1} - a) y - a_{1} b$

Now $(a - a_{1}, b - b_{1})$ also lies on this point, therefore, it should satisfy the above equation.

Thus, $(b_{1} - b) (a - a_{1}) - a b_{1} = (a_{1} - a) (b - b_{1}) - a_{1} b \Rightarrow a b_{1} - a b - a_{1} b_{1} + a_{1} b - a b_{1} = a_{1} b - a_{1} b_{1} - a b + a b_{1} - a_{1} b$ $\Rightarrow a b_{1} = a_{1} b$ .

Thus, the equation of the line becomes $(b_{1} - b) x = (a_{1} - a) y$ , which clearly passes through the origin.

The equation of the straight line which passes through $(1, 2)$ and $(- 3, 0)$ is given by

$y - 2 = \frac{0 - 2}{- 3 - 1} (x - 1) \Rightarrow 2 y = x + 3$ .

For the points to be collinear $(t - 1, 3)$ has to be on this line. Thus,

$2.3 = t - 1 + 3 => t = 4$ .

The equation of the straight line which passes through $(p, q + r)$ and $(q, r + p)$ is given by

$y - q - r = \frac{r + p - q - r}{q - p} (x - p) \Rightarrow x + y = p + q + r$ .

If the line passes through $(r, p + q)$ then it would satisfy the obtained equation of the line. Putting the point in the obtained equation we have

$r + p + q = p + q + r$ , which is true. Hence, proved.

Point dividing the line segment joining the points $(- 1, 2)$ and $(4, - 5)$ externally in the ratio $2 : 3$ is given by

$(\frac{2.4 - 3. - 1}{2 - 3}, \frac{2. - 5 - 3.2}{2 - 3}) = (- 11, 16)$ .

The equation of the line passing through $(1, 2)$ and $(- 11, 16)$ is given by

$y - 2 = \frac{16 - 2}{- 11 - 1} (x - 1) \Rightarrow 7 x + 6 y - 19 = 0$ .

The equation of $B C$ is given by $y - 1 = \frac{0 - 1}{2 - 0} (x - 0) \Rightarrow x + 2 y = 2$ .

The vertex $A$ is $(- 1, - 2)$ and median passing through it will bisect $B C$ i.e. it will pass through the point $(1, \frac{1}{2})$ .

Thus, equation of the median is given by

$y + 2 = \frac{\frac{1}{2} + 2}{1 + 1} (x + 1) \Rightarrow 5 x - 4 y - 3 = 0$ .

The mid-point of $(2, 3)$ and $(5, 4)$ is $(\frac{7}{2}, \frac{7}{2})$ . The equation of the median passing through $(1, 2)$ and $(\frac{7}{2}, \frac{7}{2})$ is given by

$y - 2 = \frac{\frac{7}{2} - 2}{\frac{7}{2} - 1} (x - 1) \Rightarrow 3 x - 5 y + 7 = 0$ .

The mid-point of $(1, 2)$ and $(2, 3)$ is $(\frac{3}{2}, \frac{5}{2})$ . The equation of the median passing through $(5, 4)$ and $(\frac{3}{2}, \frac{5}{2})$ is given by

$y - 4 = \frac{\frac{5}{2} - 4}{\frac{3}{2} - 5} (x - 5) \Rightarrow 3 x - 7 y + 13 = 0$ .

The mid-point of $(1, 2)$ and $(5, 4)$ is $(3, 3)$ . The equation of the median passing through $(2, 3)$ and $(3, 3)$ is given by

$y - 3 = \frac{3 - 3}{3 - 2} (x - 2) \Rightarrow y = 3$ .

Let the line segment joining $A (2, 3)$ and $B (- 1, 4)$ be divided by the line $x + y + 1 = 0$ in the ratio $m : n$ .

Using the section formula, the point of division $P$ is $P = (\frac{m . (- 1) + n .2}{m + n}, \frac{m .4 + n .3}{m + n})$ .

Since $P$ lies on $x + y + 1 = 0$ , substitute: $\frac{- m + 2 n}{m + n} + \frac{4 m + 3 n}{m + n} + 1 = 0$

$\Rightarrow \frac{- m + 2 n + 4 m + 3 n}{m + n} + 1 = 0 \Rightarrow \frac{3 m + 5 n}{m + n} + 1 = 0 \Rightarrow \frac{m}{n} = - \frac{3}{2}$

Hence, the line divides the segment externally in the ratio $- 3 : 2$ , i.e. $3 : 2$ externally.

Let $A (2, 3)$ and $B (4, 1)$ . Let the line through $(1, 2)$ and $(4, 3)$ divide $A B$ in the ratio $m : n$ at point $P$ .

Point $P = (\frac{m .4 + n .2}{m + n}, \frac{m .1 + n .3}{m + n})$ .

Slope of line through $(1, 2)$ and $(4, 3)$ is $\frac{3 - 2}{4 - 1} = \frac{1}{3}$ .

Equation of this line: $y - 2 = \frac{1}{3} (x - 1)$ .

Substituting $P : (\frac{m + 3 n}{m + n}) - 2 = \frac{1}{3} (\frac{4 m + 2 n}{m + n} - 1)$

Simplifying LHS: $\frac{m + 3 n - 2 m - 2 n}{m + n} = \frac{- m + n}{m + n}$

RHS: $\frac{1}{3} . \frac{4 m + 2 n - m - n}{m + n} = \frac{1}{3} . \frac{3 m + n}{m + n}$

So, $\frac{- m + n}{m + n} = \frac{3 m + n}{3 (m + n)} \Rightarrow n = 3 m \Rightarrow m : n = 1 : 3$ .

Hence, the line divides the segment internally in the ratio $1 : 3$ .

$D = (\frac{2.1 + 1. - 1}{2 + 1}, \frac{2.3 + 1. - 2}{2 + 1}) = (\frac{1}{3}, - \frac{8}{3})$

Let mid-point of $A C$ is $M$ then $M = (\frac{3}{2}, - \frac{1}{2})$ .

Equation of $B M$ is given by $y + 2 = \frac{- \frac{1}{2} + 2}{\frac{3}{2} + 1} (x + 1) \Rightarrow 5 y + 10 = 3 x + 3 \Rightarrow 3 x - 5 y = 7$

Equation of $A D$ is given by $y - 2 = \frac{- \frac{8}{3} - 2}{\frac{1}{3} - 2} (x - 2) \Rightarrow 5 y - 10 = 14 x - 28 \Rightarrow 14 x - 5 y = 18$

The point of intersection of two obtained equations is given by $(1, - \frac{4}{5})$ .

Let this point divide $B M$ in the ratio of $k : 1$ , then

$1. (k . \frac{3}{2} + 1. - 1) = (k + 1) \Rightarrow k + 1 = \frac{3}{2} . k - 1 \Rightarrow k = 4$ . Thus ratio is $4 : 1$ .

The equation of the the line can be written as $y = \sqrt{3} x + 3$ . Comparing it will $y = m x + c$ gives us $m = \sqrt{3}$ and $c = 3$ .

Thus, slope of the line $m = \sqrt{3} = \tan 60^{\circ}$ . Thus, the line makes an angle of $60^{\circ}$ with the positive direction of the $x$ -axis.

$c = 3$ tells us that the intercept on $y$ -axis is $3$ in positive direction.

Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$ .

It is given that $b = 2 a$ which makes the equation of the line $2 x + y = 2 a$ .

Since it passes through $(3, 4)$ , therefore, $2.3 + 4 = 2 a \Rightarrow a = 5$ , which makes the equation $2 x + y = 10$ .

Let the equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$ so the point on $x$ -axis where this line meets is $(a, 0)$ and on $y$ -axis it is $(0, b)$ .

Given that $(3, 4)$ divdes the line segment joining $(a, 0)$ and $(b, 0)$ in the ratio of $2 : 3$ , therefore,

$3 = (2.0 + 3. a) (2 + 3) \Rightarrow a = 5$ and $4 = (2. b + 3.0) / (2 + 3) \Rightarrow b = 10$

Thus, equation of the line is $2 x + y = 10$ .

The line $3 x + 4 y = 12$ can be written as $\frac{x}{4} + \frac{y}{3} = 1$ so the intercept of $x$ -axis is $4$ and the intercept on $y$ -axis is $3$ .

Thus, according to the question the required line makes an intercept of $8$ on $x$ -axis and $9$ on $y$ -axis. Thus, the required line is

$\frac{x}{8} + \frac{y}{9} = 1 \Rightarrow 9 x + 8 y = 72$ .

$a x + b y + c = 0$ can be written as $- \frac{x}{\frac{c}{a}} - \frac{y}{\frac{c}{b}} = 1$ . Thus, intercept on $x$ -axis is $- \frac{c}{a}$ and on $y$ -axis is $- \frac{c}{b}$ .

Let the equation of the line be $y = m x + c$ , but since the line passes through origin $c = 0$ .

Now mid-point of the intercept is given by $(- \frac{x}{2 a}, - \frac{x}{2 b})$ . Putting this point in the line

$- \frac{c}{2 b} = - \frac{m c}{2 a} \Rightarrow m = \frac{a}{b}$ , which makes the line $a x = b y$ .

Given line is $3 x + 4 y = 12 \Rightarrow \frac{x}{4} + \frac{y}{3} = 1$ . Let this line cut $x$ and $y$ axes at $A$ and $B$ respectively. Then $A = (4, 0)$ and $B = (0, 3)$ .

Let $P$ and $Q$ be the points which trisect $A B$ such that $\frac{A P}{P B} = 1 : 2$ and $\frac{A Q}{B Q} = 2 : 1$

$\Rightarrow P = (\frac{1.0 + 2.4}{3}, \frac{1.3 + 2.0}{3}) = (\frac{8}{3}, 1)$ and $Q = (\frac{2.0 + 1.4}{3}, \frac{2.3 + 1.0}{3}) = (\frac{4}{3}, 2)$

Equation of the line passing through origin and $P$ is given by $y - 0 = \frac{1 - 0}{\frac{8}{3} - 0} (x - 0) \Rightarrow 3 x - 8 y = 0$ .

Equation of the line passing through origin and $Q$ is given by $y - 0 = \frac{2 - 0}{\frac{4}{3} - 0} (x - 0) \Rightarrow 3 x - 2 y = 0$ .

Let the line be $\frac{x}{a} + \frac{y}{b} = 1$ , which will cut intercepts $a$ and $b$ . According to question

$\frac{1}{a} + \frac{1}{b} = k$ , where $k$ is a constant. Thus, $\frac{\frac{1}{k}}{a} + \frac{\frac{1}{k}}{b} = 1$ , which passes through the point $(\frac{1}{k}, \frac{1}{k})$ .

Let $A B C$ be a right angles isosceles triangle in which $A B = A C$ . We take $A$ as the origin and $A B$ and $A C$ as $x$ and $y$ axes respectively. Let $A B = A C = a$ .

Also, let $A P = h, A Q = k$ . The equation of the line $P Q$ is $\frac{x}{b} + \frac{y}{k} = 1$

Given that $B P . C Q = A B^{2} \Rightarrow (h - a) (k - a) = a^{2} \Rightarrow \frac{a}{h} + \frac{a}{k} = 1$ , which shows that $P Q$ passes through the point $(a, a)$ .

Figure 9.5.

Given that $P = (α, β)$ and the equation of the line $\frac{x}{a} + \frac{y}{b} = 1$ .

The line will cut the axes at $(a, 0)$ and $(0, b)$ . Given that $Δ O A B = \frac{1}{2} | a b | = S \Rightarrow 2 S = a b$ , where $O$ is the origin.

Since the line passes through $P$ , therefore $\frac{α}{a} + \frac{β}{b} = 1 \Rightarrow \frac{α}{a} + \frac{a β}{2 S} = 1$

$\Rightarrow a^{2} β - 2 a S + 2 α S = 0$ , which is a quadratic equation in $a$ . However, $a$ is real, therefore $D = 4 S^{2} - 8 α β S \geq 0 \Rightarrow S \geq 2 α β$

Thus, the least value of $S$ is $2 α β$ .

The equation of the line will be given by $x \cos 75^{\circ} + y \sin 75^{\circ} = 3 \sqrt{2}$

Now $\cos 75^{\circ} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$ and $\sin 75^{\circ} = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

So the equation of the line is $(\sqrt{3} - 1) x + (\sqrt{3} + 1) y = 12$ .

Slope is given as $\frac{5}{12}$ so if $\tan θ = \frac{5}{12}$ then $θ$ can lie in first or third quadrant. Thus, $\cos θ = \pm \frac{12}{13}$ and $\sin θ = \pm \frac{5}{13}$ .

Equation of the line will be $x . \frac{12}{13} + y . \frac{5}{13} = 2 \Rightarrow 12 x + 5 y - 26 = 0$

or $x . (- \frac{12}{13}) + y . (- \frac{5}{13}) = 2 \Rightarrow 12 x + 5 y + 26 = 0$ .

We can treat this place as origin, east direction as $x$ -axis and north direction as $y$ -axis. Then the angle made by perpendicular from the place to the line will be $45^{\circ}$ as the direction of the canal is north-east.

Thus, equation for this canal would be $\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = \frac{9}{2} \Rightarrow x + y = \frac{9}{\sqrt{2}}$ .

The coordinate of the village is given by $(3, 4)$ ; putting this in the equation for the canal gives us

$3 + 4 = \frac{9}{\sqrt{2}}$ , which is false. Hence, the village does not lie on the canal.

Let the reuired line be $A B$ and $O L$ is perpendicular from the origin $O$ to $A B$ . According to the question $O L$ makes an angle of $30^{\circ}$ with $y$ -axis i.e. it will make an angle of $60^{\circ}$ with $x$ -axis.

Let $O L = p$ , so the equation of the line will be $x \cos 60^{\circ} + y \sin 60^{\circ} = p \Rightarrow x + \sqrt{3} y = 2 p$ .

Intercept on $x$ -axis is $2 p$ and intercept on $y$ -axis is $\frac{2 p}{\sqrt{3}}$ .

Also given that $Δ O A B = 96 \sqrt{3} = \frac{1}{2} . O A . O B \Rightarrow 96 \sqrt{3} = \frac{1}{2} .2 p . \frac{2 p}{\sqrt{3}} \Rightarrow p = 12$

Thus, equation of the line is $x + \sqrt{3} y = 24$ .

Figure 9.6.

Given $O C = 2, ∠ A B C = 90^{\circ}$ and $A B = B C$

$∴ ∠ B C A = ∠ B A C = 45^{\circ}, O B = O C = 2$ , and $B C = \sqrt{2^{2} + 2^{2}} = 2 \sqrt{2}$

Let $O M$ be perpendicular to $D E$ . $∵ O C = 2, O B = 2$ , therefore, equation of $B C$ will be

$\frac{x}{2} + \frac{y}{2} = 1 \Rightarrow x + y = 2$

Also, equation of $A B$ will be $\frac{x}{- 2} + \frac{y}{2} = 1 \Rightarrow x - y = - 2$

$∵ ∠ L A C = 45^{\circ}$ and $O M = O L + L M = O C \cos 45^{\circ} + L M = O C . \frac{1}{\sqrt{2}} + B C = 3 \sqrt{2}$

Thus, equation of $D E$ is $x \cos 45^{\circ} + y \sin 45^{\circ} = 3 \sqrt{2} \Rightarrow x + y = 6$ .

Given equation is $\sqrt{3} x + y = 8 \Rightarrow \frac{x}{\frac{8}{\sqrt{3}}} + \frac{y}{8} = 1$ , which will meet $x$ and $y$ axes at $(\frac{8}{\sqrt{3}}, 0)$ and $(0, 8)$ .

The equation can be rewritten as $\sqrt{3} 2 x + \frac{y}{2} = 4 \Rightarrow x \cos 30^{\circ} + y \sin 30^{\circ} = 4$ , which is the equation in normal form.

The length of perpendicular on this line from origin is $4$ and it makes an angle of $30^{\circ}$ with the $x$ -axis.

Let the equation of a line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$ .

Since it passes through $(3, 2)$ , we have: $\frac{3}{a} + \frac{2}{b} = 1$ .

Given $a - b = 2$ , so $a = b + 2$ .

Substitute into the first equation: $\frac{3}{b + 2} + \frac{2}{b} = 1$ .

Multiplying by $b (b + 2) : 3 b + 2 (b + 2) = b (b + 2), b^{2} - 3 b - 4 = 0$

Solving: $b = \frac{3 \pm 5}{2}$ . So, $b = 4$ or $b = - 1$ .

Then $a = b + 2$ gives: If $b = 4$ , then $a = 6$ . If $b = - 1$ , then $a = 1$ .

Thus, the required lines are: $\frac{x}{6} + \frac{y}{4} = 1$ and $x - y = 1$ .

Let $a$ be $(a, 0)$ and $B$ be $(0, b)$ , then the equation of line will be given by $\frac{x}{a} + \frac{y}{b} = 1$ .

Since it passes through $P (1, - 7)$ , therefore, $\frac{1}{a} - \frac{7}{b} = 1 \Rightarrow \frac{1}{a} = \frac{7 + b}{b} \Rightarrow a = \frac{b}{7 + b}$

Also given that $4 A P = 4 B P \Rightarrow 16 [(a - 1)^{2} + 7^{2}] = 9 [1^{2} + (- 7 - b)^{2}]$

$\Rightarrow 16 (a - 1)^{2} + 784 = 9 + (b + 7)^{2}$

Putting the value of $a$ from above we get $b = - \frac{49}{3}$ and $a = \frac{7}{4}$

Thus, equation of the line is $28 x - 3 y = 49$ .

Let $A$ be $(a, 0)$ abd $B$ be $(0, b)$ , then the equation of line will be given by $\frac{x}{a} + \frac{y}{b} = 1$ .

Since it passes through $P (2, 6)$ , therefore, $\frac{2}{a} + \frac{6}{b} = 1 \Rightarrow \frac{2}{a} = \frac{b - 6}{b} \Rightarrow a = \frac{2 b}{b - 6}$

Also given that $3 A P = 2 B P \Rightarrow 9 [(a - 2)^{2} + 6^{2}] = 4 [(- 2)^{2} + (b - 6)^{2}]$

Putting $a = \frac{2 b}{b - 6}$ and solving gives us $a = \frac{10}{3}$ and $b = 15$

Thus, equation of the line is $9 x + 2 y = 30$ .

Given line is $3 x + 4 y = 6 \Rightarrow \frac{x}{2} + \frac{y}{\frac{3}{2}} = 1$ . Thus, intercepts on axes are $2$ and $\frac{3}{2}$ respectively.

Double of these intercepts is $4$ and $3$ . Thus, equation of line which makes these intercepts is

$\frac{x}{4} + \frac{y}{3} = 1 \Rightarrow 3 x + 4 y = 12$ .

Given line is $3 x - 5 y = 15 \Rightarrow \frac{x}{5} - \frac{y}{3} = 1$ . Thus, points of interception are $(5, 0)$ and $(0, - 3)$ . Midpoint of intercepted portion will be $(\frac{5}{2}, - \frac{3}{2})$ .

The required line also passes through $(2, 1)$ , hence in two-point form equation of the line will be

$y - 1 = \frac{- \frac{3}{2} - 1}{\frac{5}{2} - 1} (x - 2) \Rightarrow 5 x + y = 11$ .

The given line is $2 x + 3 y = 6 \Rightarrow \frac{x}{3} + \frac{y}{2} = 1$ , thus points of interception are $(3, 0)$ and $(0, 2)$ .

Let the points be $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ which divide the intercepted points in the ratio of $2 : 1$ and $1 : 2$ respectively.

Thus, $P = (\frac{2.0 + 1.3}{3}, \frac{2.2 + 1.0}{3}) = (1, \frac{4}{3})$ and $Q = (\frac{1.0 + 2.3}{3}, \frac{1.2 + 2.0}{3}) = (2, \frac{2}{3})$ .

Since these lines also pass through origin so the equations are given by $y = 4 / 3 x \Rightarrow 4 x - 3 y = 0$ and $y = \frac{\frac{2}{3}}{2 x} \Rightarrow x - 3 y = 0$ .

Equation of the line in two-point form is given by $y - 1 = \frac{4 - 1}{11 - 5} (x - 5) \Rightarrow 2 y - 2 = x - 5 \Rightarrow x - 2 y - 3 = 0$

Putting $(1, - 1)$ in the obtained equation for the line $1 - 2 (- 1) - 3 = 0$ , which is true, so all points lie on the line $x - 2 y = 3 \Rightarrow \frac{x}{3} + \frac{y}{- \frac{3}{2}} = 1$ .

Thus, intercepts on the axes are $(3, 0)$ and $(0, - \frac{3}{2})$ and intercepts between the axes is $\sqrt{3^{2} + ({\frac{3}{2}}^{2})} = \frac{3 \sqrt{5}}{2}$ .

Equation of the line in two-point form is given by $y + 3 = \frac{5 + 3}{4 - 1} (x - 1) \Rightarrow 3 y + 9 = 8 x - 8 \Rightarrow 8 x - 3 y = 17 \Rightarrow \frac{x}{\frac{17}{8}} + \frac{y}{- \frac{17}{3}} = 1$ .

Thus, intercepts on the axes are $\frac{17}{8}$ and $- \frac{17}{3}$ respectively.

There are two possibilities as shown in the diagram because length is a scalar quantity.

Figure 9.7.

Since the line makes an angle of $150^{\circ}$ with positive direction of $y$ -axis so it will make an angle of $120^{\circ}$ with positive direction of $x$ -axis.

Thus, angle made by perpendicular with $x$ -axis would be $30^{\circ}$ or $210^{\circ}$ with positive direction of $x$ -axis.

Thus, equation of the line is $x \cos 30^{\circ} + y \sin 30^{\circ} = 7$ and $x \cos 210^{\circ} + y \sin 210^{\circ} = 7$ .

Thus, lines are given by $\sqrt{3} x + y = \pm 7$ .

Since the perpendicular makes an angle with positive direction of $y$ -axis with $30^{\circ}$ it will make an angle of $60^{\circ}$ with positive direction of $x$ -axis. Also, given that length of the perpendicular from origin is $2$ . Therefore, the equation in normal form is given by

$x \cos 60^{\circ} + y \sin 60^{\circ} = 2 \Rightarrow x + \sqrt{3} y = 4$ .

The equation of the line in normal form is given by

$x \cos 60^{\circ} + y \sin 60^{\circ} = 5 \Rightarrow x + \sqrt{3} y = 10$ .

Given that $\tan θ = \frac{3}{4}$ where $t h e t a$ is the angle made by the perpendicular with the positive direction of $x$ -axis. Thus, $\tan θ$ can also be $\frac{- 3}{- 4}$ i.e. in third quadrant.

$\Rightarrow \cos θ = \pm \frac{4}{5}, \sin θ = \pm \frac{3}{5}$ and the equation in normal form will be

$x \cos θ + y \sin θ = 6 \Rightarrow 4 x + 3 y = \pm 30$ .

The equation of the line joining the points $(1, 2)$ and $(- 3, 1)$ is given by

$y - 2 = \frac{1 - 2}{- 3 - 1} (x - 1) \Rightarrow x - 4 y + 7 = 0$ , which can be written as $\frac{x}{7} + \frac{y}{- \frac{7}{4}} = 1$ .

Thus, intercepts on axes are $7$ and $- \frac{7}{4}$ .

$\cos θ = \frac{7}{\sqrt{7^{2} + \frac{7^{2}}{4^{2}}}} \Rightarrow p = 7 / \sqrt{17}$ .

Let $P = (3, 2)$ and let the required line make an angle $θ$ with the positive direction of $x$ -axis.

Given $\tan θ = \frac{3}{4}$ .

So the equation of the line is $y - 2 = \frac{3}{4} (x - 3) \Rightarrow 3 x - 4 y - 1 = 0$ .

Coordinates of the points which are at a distance of $5$ units from $P$ are $(3 \pm 5 \cos θ, 2 \pm 5 \sin θ) \Rightarrow (3 \pm 4, 2 \pm 3) \Rightarrow (7, 5)$ or $(- 1, - 1)$ .

Let $P = (1, 2)$ . Let $A B$ be the given line $x + y = 4$ .

Let the line through $P$ makes an angle $θ$ with the $x$ -axis cuts the line $A B$ at $Q$ and $R$ at a distance $\sqrt{\frac{2}{3}}$ from $P$ . Then

$Q = (1 + \sqrt{\frac{2}{3}} \cos θ, 2 + \sqrt{\frac{2}{3}} \sin θ)$

Since $Q$ lies on the line $A B$ therefore $1 + \sqrt{\frac{2}{3}} \cos θ + 2 + \sqrt{\frac{2}{3}} \sin θ = 4$

$\Rightarrow \cos θ + \sin θ = \sqrt{\frac{3}{2}} => \frac{1}{\sqrt{2}} \cos θ + \frac{1}{\sqrt{2}} \sin θ = \frac{\sqrt{3}}{2}$

$\Rightarrow c o s (θ - 45^{\circ}) = c o s 30^{\circ} \Rightarrow θ - 45^{\circ} = 2 n π \pm 30^{\circ}$

$=> θ = 15^{\circ}, 75^{\circ}$ .

Figure 9.8.

Given line is $\sqrt{3} x - 4 y + 8 = 0$ and $P = (\sqrt{3}, 2)$ . Let the line through $P$ making an angle of $\frac{π}{6}$ with the $x$ -axis meet the libe at $Q$ . Let $P Q = r$ , then

$Q = (\sqrt{3} + r \cos \frac{π}{6}, 2 + r \sin \frac{π}{6}) = (\sqrt{3} + \frac{\sqrt{3}}{2} r, 2 + \frac{r}{2})$

However, $Q$ lies on the given line, therefore,

$\sqrt{3} (\sqrt{3} + \frac{\sqrt{3}}{2}) - 4 (2 + \frac{r}{2}) + 8 = 0 \Rightarrow r = 6$ .

Let $P = (- 2, 3)$ . We know that the coordinates of points on the line making an angle $θ$ with the positive direction of $x$ -axis at a distance $r$ from a point $(x_{1}, y_{1})$ are $(x_{1} \pm r \cos θ, y_{1} \pm r \sin θ)$ .

Thus, required coordinates are $(- 2 \pm 4 \sqrt{2} \cos 45^{\circ}, 3 \pm 4 \sqrt{2} \sin 45^{\circ})$ i.e. $(2, 7)$ and $(- 6, - 1)$ .

Given $A = (2, 0)$ and $B = (3, 1)$ . Slope of the line $A B = \frac{0 - 1}{2 - 3} = 1 = \tan 45^{\circ}$ .

Thus, slope of the line $A C = 45^{\circ} + 15^{\circ} = 60^{\circ}$

Therefore, equation of the line $A C$ is $y - 0 = \tan 60^{\circ} (x - 2) \Rightarrow \sqrt{3} x - y = 2 \sqrt{3}$ .

$A C = A B = \sqrt{(3 - 2)^{2} + (1 - 0)^{2}} = \sqrt{2}$

Thus, $C = (2 + \sqrt{2} \cos 60^{\circ}, 0 + \sqrt{2} \sin 60^{\circ}) = (2 + \frac{1}{\sqrt{2}}, \sqrt{\frac{3}{2}})$ .

Let $A = (1, 1)$ and $C = (- 2, - 1)$ then $H = (- \frac{1}{2}, 0)$ , which is mid-point of $A C$ and $B D$ .

Slope of $A C = \frac{1 + 1}{1 + 2} = \frac{2}{3} = \tan θ$ , therefore, slope of $B C = - \frac{3}{2}$ because diagonals of a square are perpendicular to each other.

Thus, $t h e t a$ is an obtuse angle. $∴ \cos θ = - \frac{2}{13}$ and $\sin θ = \frac{3}{\sqrt{13}}$

Also, $A C = \sqrt{13}$ , therefore, $D H = \frac{\sqrt{13}}{2}$

Thus, coordinates of $B$ and $D$ are $(- \frac{1}{2} \pm \frac{\sqrt{13}}{2} \cos θ, 0 \pm \frac{\sqrt{13}}{2} \sin θ)$ i.e. $(- \frac{3}{2}, \frac{3}{2})$ and $(\frac{1}{2}, - \frac{3}{2})$ .

Figure 9.9.

Let the line through $A$ making an angle $θ$ with the positive direction at $x$ -axis. Let $A B = r_{1}, A C = r_{2}$ and $A D = r_{3}$ .

$B = (k + 1 + r_{1} \cos θ, 2 k + r_{1} \sin θ)$ . Since $B$ lies on $7 x + y - 16 = 0$ , therefore,

$7 (k + 1 + r_{1} \cos θ) + 2 k + r_{2} \sin θ - 16 = 0 \Rightarrow r_{1} = \frac{9 (1 - k)}{7 \cos θ + \sin θ}$

Also, $C = (k + 1 + r_{2} \cos θ, 2 k + r_{2} \sin θ)$ . Since $C$ lies on the line $5 x - y - 8 = 0$ , therefore,

$5 (k + 1 + r_{2} \cos θ) - (2 k + r_{2} \sin θ) - 8 = 0 \Rightarrow r_{2} = \frac{3 (1 - k)}{5 \cos θ - \sin θ}$

Again $D = (k + 1 + r_{3} \cos θ, 2 k + r_{3} \sin θ)$ and $D$ lies on the line $x - 5 y + 8 = 0$ , therefore,

$k + 1 + r_{3} \cos θ - 5 (2 k + r_{3} \sin θ) + 8 = 0 \Rightarrow r_{3} = \frac{9 (1 - k)}{5 \sin θ - \cos θ}$

$\frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{2 (7 \cos θ + \sin θ)}{9 (1 - k)} = \frac{2}{r_{1}}$ .

Hence, $r_{2}, r_{1}, r_{3}$ are in H.P.

Let $A B C D$ be the square whose center is $O$ . Now $A O = \sqrt{5}$ and slope of $A O = \frac{1 - 0}{2 - 0} = \frac{1}{2} = \tan θ$

$∴ \cos θ = \frac{2}{\sqrt{5}}$ and $\sin θ = \frac{1}{\sqrt{5}}$

$∴$ Coordinates of the points of $A C$ which are at a distance $\sqrt{5}$ from $O$ will be $(0 \pm \sqrt{5} \cos θ, \pm \sqrt{5} \sin θ) = (\pm 2, \pm 1)$

i.e. $(2, 1)$ and $(- 2, - 1)$ . Thus, $C = (- 2, - 1)$ .

But $B D ⊥ A C$ . So slope of $B D = - 2 = \tan α$ (say)

$∴ \frac{π}{2} < α < π$ or $\frac{3 π}{2} < α < 2 π$

$∴ \cos α = - \frac{1}{\sqrt{5}}$ and $\sin α = \frac{2}{\sqrt{5}}$ or $\cos α = \frac{1}{\sqrt{5}}$ and $\sin α = - \frac{2}{\sqrt{5}}$

Since $B$ and $D$ are on $B D$ at a distance $\sqrt{5}$ from $O$ , their coordinates(in some order) will be

$(0 \pm \sqrt{5} \cos α, 0 \pm \sqrt{5} \sin α)$ i.e. $(\pm 1, \pm 2)$ .

Let $A D$ be the internal bisector of $∠ B A C$ then $\frac{B D}{D C} = \frac{A B}{A C} = \frac{c}{b}$

Thus, $D = (\frac{b x_{2} + c x_{3}}{b + c}, \frac{b y_{2} + c y_{3}}{b + c})$

Let the equation of the line $A D$ be $l x + m y + n = 0$ , then we observe that $A$ and $D$ lie on this line. Therefore

$l x_{1} + m y_{1} + n = 0$ and $l (\frac{b x_{2} + c x_{3}}{b + c}) + m (\frac{b y_{2} + c y_{3}}{b + c}) + n = 0$

Eliminating $l, m, n$ gives us

$| \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ \frac{b x_{2} + c x_{3}}{b + c} & \frac{b y_{2} + c y_{3}}{b + c} & 1 \end{matrix} | = 0 \Rightarrow b | \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{matrix} | + c | \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} | = 0$ .

The required points are $(1 \pm 6 \cos 60^{\circ}, 1 \pm 6 \sin 60^{\circ})$ i.e. $(4, 1 + 3 \sqrt{3})$ and $(- 2, 1 - 3 \sqrt{3})$ .

The equation of the line passing through $(- 1, 3)$ and slope $1$ is given by $y - 3 = x + 1 \Rightarrow x - y + 4 = 0$ .

Putting $x = y - 4$ in the given equation $2 y - 8 + y = 3 \Rightarrow y = \frac{11}{3}$ and $x = - \frac{1}{3}$ .

Distance between $(- 1, 3)$ and $(- \frac{1}{3}, \frac{11}{3})$ is $\sqrt{{(- \frac{1}{3} + 1)}^{2} + {(\frac{11}{3} - 3)}^{2}} = \frac{2 \sqrt{2}}{3}$ .

Let the line through $P (x_{1}, y_{1})$ inclined at angle $θ$ with the $x$ -axis have slope $\tan θ$ . Its equation is $y - y_{1} = \tan θ (x - x_{1})$ .

Rewriting, $\tan θ x - y + (y_{1} - x_{1} \tan θ) = 0$ .

The given line is $a x + b y + c = 0$ .

If $Q$ is the intersection point, the distance $P Q$ measured along the direction making angle $θ$ with the $x$ -axis is $P Q = | \frac{a x_{1} + b y_{1} + c}{a \cos θ + b \sin θ} |$ .

Give that the line makes an angle of $30^{\circ}$ with positive direction of $x$ -axis and rotated $15^{\circ}$ in anticlockwise direction so the line will now make $45^{\circ}$ with the positive direction of $x$ -axis.

Thus, slope of the line is $\tan 45^{\circ} = 1$ . Also, the line passes through $(2, 0)$ so the equation of line is

$y - 0 = 1. (x - 2) \Rightarrow x - y - 2 = 0$ .

Given the line $2 x - y = 5$ . Substitute $y = x$ into the equation: $2 x - x = 5 \Rightarrow x = 5$ .

So the point of rotation is $(5, 5)$ . Slope $m = 2$ .

After rotation by $45^{\circ}$ , the new slope is: $m^{'} = \tan (\tan^{- 1} 2 + 45^{\circ})$ .

Using the identity: $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ , we get:

$m^{'} = \frac{2 + 1}{1 - 2.1} = - 3$ .

Using point-slope form: $y - 5 = - 3 (x - 5) \Rightarrow y = - 3 x + 20$ .

The given line is $x + 2 y = 4$ . The line is translated by $3$ units in the direction of increasing $x$ . So replace $x$ with $x - 3$ :

$(x - 3) + 2 y = 4 \Rightarrow x + 2 y = 7$ .

Now the shifted line cuts the $x$ -axis at ( $y = 0$ ): $x = 7$ . So the pivot point is $(7, 0)$ .

From $x + 2 y = 7 : y = - \frac{1}{2} x + \frac{7}{2}$ , so slope $m = - \frac{1}{2}$ .

Angle of inclination $θ$ satisfies $\tan θ = - \frac{1}{2}$ .

After clockwise rotation by $30^{c} i r c$ , new angle is $θ - 30^{\circ}$ .

New slope: $m^{'} = \tan (θ - 30^{\circ})$

$\tan (θ - 30^{\circ}) = \frac{\tan θ - t a n 30^{\circ}}{1 + \tan θ \tan 30^{\circ}}$

$\Rightarrow m^{'} = \frac{- \frac{1}{2} - \frac{1}{\sqrt{3}}}{1 + (- \frac{1}{2}) (\frac{1}{\sqrt{3}})} \Rightarrow - \frac{1}{2} - \frac{1}{\sqrt{3}} = \frac{- \sqrt{3} - 2}{2 \sqrt{3}}$ .

$\Rightarrow 1 - \frac{1}{2 \sqrt{3}} = \frac{2 \sqrt{3} - 1}{2 \sqrt{3}}$ . So, $m^{'} = - \frac{\sqrt{3} + 2}{2 \sqrt{3} - 1}$ .

Using point-slope form with point $(7, 0) : y - 0 = m^{'} (x - 7)$

$\Rightarrow y = - \frac{\sqrt{3} + 2}{2 \sqrt{3} - 1} (x - 7)$ .

Let the regular hexagon be $A B C D E F$ with side length $a$ , and $A$ as origin $(0, 0)$ . Given $A B$ lies along the $x$ -axis and $A E$ along the $y$ -axis.

Since $A B = a$ along $x$ -axis: $B = (a, 0)$

Since $A E = a$ along $y$ -axis: $E = (0, a)$

In a regular hexagon, each interior angle is $120^{\circ}$ , so directions change by $60^{\circ}$ .

Direction $B C$ makes $60^{\circ}$ with $A B : C = B + (a \cos 60^{\circ}, a \sin 60^{\circ}) = (a + \frac{a}{2}, a \frac{\sqrt{3}}{2}) = (\frac{3 a}{2}, \frac{\sqrt{3}}{2})$ .

$D = C + (a \cos 120^{\circ}, a \sin 120^{\circ}) = (\frac{3 a}{2} - \frac{a}{2}, \frac{\sqrt{3} a}{2} + \frac{\sqrt{3} a}{2}) = (a, \sqrt{3} a)$

$F = A + (a \cos (- 60^{\circ}), a \sin (- 60^{\circ})) = (\frac{a}{2}, - \frac{\sqrt{3} a}{2})$

Equation of $A C : A (0, 0)$ and $C (\frac{3 a}{2}, \frac{\sqrt{3} a}{2})$ , $m = (\frac{\frac{\sqrt{3} a}{2}}{\frac{3 a}{2}}) = \frac{\sqrt{3}}{3}$ $\Rightarrow A C : y = \frac{\sqrt{3}}{2}$

For $A F : A (0, 0)$ and $F (\frac{a}{2}, - \frac{\sqrt{3} a}{2})$ , $m = \frac{- \frac{\sqrt{3} a}{2}}{\frac{a}{2}} = - \sqrt{3}$

Equation of $A F : y = - \sqrt{3} x$

For $B E : B (a, 0)$ and $E (0, a), m = \frac{a - 0}{0 - a} = - 1$

$y - 0 = - 1 (x - a) \Rightarrow y = - x + a$ .

Let the place be the origin $O (0, 0)$ . The road is at a perpendicular distance $5 \sqrt{2}$ from $O$ , and the shortest distance is in the $N - E$ direction, i.e., along a line making $45^{\circ}$ with the axes.

So the normal to the road has slope $1$ , hence the road has slope $- 1$ .

Thus, equation of the road is of form: $y = - x + c$

Distance from origin to this line: $\frac{| c |}{s q r t 1^{2} + (- 1)^{2}} = \frac{| c |}{\sqrt{2}}$

Given distance is $5 \sqrt{2} : \frac{| c |}{\sqrt{2}} = 5 \sqrt{2} \Rightarrow | c | = 10$

Since direction is $N - E$ , take positive value: $\Rightarrow c = 10$

So road equation: $y = - x + 10$

(i) Check point $(6, 4) : 4 = - 6 + 10 \Rightarrow 4 = 4$ . So, village lies on the road.

(ii) Check point $(4, 3) : 3 = - 4 + 10 \Rightarrow 3 = 6$ So, village does not lie on the road.

Given line: $x - y + 1 = 0 \Rightarrow y = x + 1$ so slope $m = 1$

Point of rotation (on $y$ -axis): $x = 0 \Rightarrow y = 1$ so $A (0, 1)$

Angle of inclination: $\tan θ = 1 \Rightarrow θ = 45^{\circ}$

After clockwise rotation by $75^{\circ}$ : new angle $= 45^{\circ} - 75^{c} i r c = - 30^{\circ}$

New slope: $m^{'} = \tan (- 30^{\circ}) = - \frac{1}{\sqrt{3}}$

Equation using point-slope form at $A (0, 1) : y - 1 = - \frac{1}{\sqrt{3}} (x - 0) \Rightarrow y = 1 - \frac{x}{\sqrt{3}}$ .

The diagram is smae as problem 77. $O C$ is $2$ units therefore $O B = 2$ units. From the diagram we see that extended $B E$ makes an angle of $45^{\circ}$ with $x$ -axis.

Slope: $m = \tan 45^{\circ} = 1$ . Equation is intercept form: $y = m x + c \Rightarrow y = x + 2$ .

$C D$ will have same slope but passes through $(0, 2)$ . Equation in slope-point form: $y - 0 = 1. (x - 2) \Rightarrow x - y = 2$ .

The midpoint is $(\frac{3 + 1}{2}, \frac{- 1 + 1}{2}) = (2, 0)$

Slope: $m = \frac{1 - (- 1)}{1 - 3} = - 1$

A line perpendicular to this will have slope equal to the negative reciprocal of $- 1$ , which is $1$ .

Let the required point be $(x, y)$ . Since it lies on the perpendicular line passing through $(2, 0)$ , its equation is:

$y - 0 = 1 (x - 2) => y = x - 2$

Also, the distance from $(2, 0)$ to $(x, y)$ is $2$ , so: $\sqrt{(x - 2)^{2} + (y - 0)^{2}} = 2$

$\Rightarrow (x - 2)^{2} + y^{2} = 4$ . Substitute $y = x - 2$ :

$(x - 2)^{2} + (x - 2)^{2} = 4 \Rightarrow 2 (x - 2)^{2} = 4 \Rightarrow (x - 2)^{2} = 2$

$x - 2 = \pm \sqrt{2} \Rightarrow x = 2 \pm \sqrt{2}$

Then $y = x - 2$ gives: $y = \pm \sqrt{2}$

Since the shift is in the sense of increasing $y$ , we take the positive value:

$x = 2 + \sqrt{2}, y = \sqrt{2}$

The given line is $2 x = y$ , i.e. $y = 2 x$ .

So, the slope of the line is $2$ . A direction along this line can be taken as $(1, 2)$ .

Now, its length is $\sqrt{1^{2} + 2^{2}} = \sqrt{5}$

So, the unit direction along the line is $(\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}})$

Since the translation is in the first quadrant, both coordinates increase.

Add this to the point $(1, 1)$ :

New point $= (1 + \frac{1}{\sqrt{5}}, 1 + \frac{2}{\sqrt{5}})$ .

We are given $A (2, - 1)$ and the line $x - y = 3$ .

Let the required point be $A^{'} (x, y)$ . Since the translation is parallel to the line, the slope of the line joining $A$ and $A^{'}$ must be equal to the slope of $x - y = 3$ .

Rewrite the line: $y = x - 3$ , so slope $= 1$ .

Hence, $\frac{y - (- 1))}{x - 2} = 1 \Rightarrow y + 1 = x - 2 \Rightarrow y = x - 3$ .

Now use the distance condition: Distance between $A (2, - 1)$ and $A^{'} (x, y)$ is $4$ .

So, $(x - 2)^{2} + (y + 1)^{2} = 16$ .

Substitute $y = x - 3 : (x - 2)^{2} + ((x - 3) + 1)^{2} = 16 \Rightarrow (x - 2)^{2} + (x - 2)^{2} = 16 \Rightarrow 2 (x - 2)^{2} = 16 \Rightarrow x - 2 = \pm \sqrt{8} = \pm 2 \sqrt{2}$ .

So, $x = 2 \pm 2 \sqrt{2}$ . Then, $y = x - 3 = - 1 \pm 2 \sqrt{2}$ .

Thus the two possible points are: $(2 + 2 \sqrt{2}, - 1 + 2 \sqrt{2})$ and $(2 - 2 \sqrt{2}, - 1 - 2 \sqrt{2})$ .

Both particles start from $A (2, - 1)$ . First particle moves along the line $x + y = 1$ . Rewrite: $y = 1 - x$ , so slope $= - 1$ .

Let its new position be $(x_{1}, y_{1})$ . Since it moves towards increasing $y$ , we take direction where $y$ increases.

Using slope condition: $\frac{y_{1} - (- 1)}{x_{1} - 2} = - 1 \Rightarrow y_{1} + 1 = - (x_{1} - 2) \Rightarrow y_{1} = - x_{1} + 1$ .

Distance moved is $2$ , so $(x_{1} - 2)^{2} + (y_{1} + 1)^{2} = 4$ .

Substitute $y_{1} = - x_{1} + 1 : (x_{1} - 2)^{2} + (- x_{1} + 1 + 1)^{2} = 4 \Rightarrow (x_{1} - 2)^{2} + (- x_{1} + 2)^{2} = 4 \Rightarrow (x_{1} - 2)^{2} + (x_{1} - 2)^{2} = 4 \Rightarrow 2 (x_{1} - 2)^{2} = 4 \Rightarrow (x_{1} - 2)^{2} = 2 => x_{1} - 2 = \pm \sqrt{2}$ .

So, $x_{1} = 2 \pm \sqrt{2}$ . Then $y_{1} = - x_{1} + 1 = - 1 \pm \sqrt{2}$ .

Since $y$ must increase from $- 1$ m we take $y_{1} = - 1 + \sqrt{2}$ , hence $x_{1} = 2 - \sqrt{2}$ .

So first particle's position: $(2 - \sqrt{2}, - 1 + \sqrt{2})$ .

Second particle moves along $x - 2 y = 4$ . Rewrite: $y = \frac{x - 4}{2}$ , so slope $= \frac{1}{2}$ .

Let position be $(x_{2}, y_{2})$ .

Slope: $\frac{y_{2} - (- 1)}{x_{2} - 2} = \frac{1}{2} \Rightarrow y_{2} + 1 = \frac{x_{2} - 2}{2} \Rightarrow y_{2} = \frac{x_{2}}{2} - 2$ .

Distance moved is $5$ , so $(x_{2} - 2)^{2} + (y_{2} + 1)^{2} = 25$ .

Substitute $y_{2} = \frac{x_{2}}{2} - 2 : (x_{2} - 2)^{2} + {(\frac{x_{2}}{2} - 2 + 1)}^{2} = 25 \Rightarrow (x_{2} - 2)^{2} + {(\frac{x_{2}}{2} - 1)}^{2} = 25$ .

$(x_{2} - 2)^{2} = x_{2}^{2} - 4 x_{2} + 4 \Rightarrow {(\frac{x_{2}}{2} - 1)}^{2} = \frac{x_{2}^{2}}{4} - x_{2} + 1$ .

So, $x_{2}^{2} - 4 x_{2} + 4 + \frac{x_{2}^{2}}{4} - x_{2} + 1 = 25 \Rightarrow \frac{5}{4} x_{2}^{2} - 5 x_{2} + 5 = 25 \Rightarrow \frac{5}{4} x_{2}^{2} - 5 x_{2} - 20 = 0$ multiply by $4 : 5 x_{2}^{2} - 20 x_{2} - 80 = 0 \Rightarrow x_{2}^{2} - 4 x_{2} - 16 = 0$ .

$\Rightarrow x_{2} = \frac{4 \pm \sqrt{16 + 64}}{2} = \frac{4 \pm \sqrt{80}}{2} = 2 \pm 2 \pm \sqrt{5}$ . Then $y_{2} = \frac{x_{2}}{2} - 2 = - 1 \pm \sqrt{5}$ .

Since $y$ increases from $- 1$ , take $y_{2} = - 1 + \sqrt{5}$ , so $x_{2} = 2 + 2 \sqrt{5}$ .

Thus, second particle's position: $(2 + 2 \sqrt{5}, - 1 + \sqrt{5})$ .

Distance between the two new positions:

${[(2 + 2 \sqrt{5}) - (2 - \sqrt{2})]}^{2} + {[(- 1 + \sqrt{5}) - (- 1 + \sqrt{2})]}^{2} = \sqrt{29 + 2 \sqrt{10}}$ .

We are given fixed point $A (4, - 1)$ and the other end $B (1, 2)$ . Let the new position of $B$ after stretching be $B^{'} (x, y)$ .

Since the string remains straight, points $A, B$ , and $B^{'}$ are collinear.

Slope of $A B : \frac{2 - (- 1)}{1 - 4} = - 1$ .

So equation of line through $A : \frac{y - (- 1)}{x - 4} = - 1 \Rightarrow y + 1 = - (x - 4) \Rightarrow y = - x + 3$ .

Length of $A B : A B^{2} = (1 - 4)^{2} + (2 + 1)^{2} = (- 3)^{2} + 3^{2} = 9 + 9 = 18$ .

Since the string is stretched to triple its length, $A B^{'} = 3 \sqrt{18}$ .

Thus, $(x - 4)^{2} + (y + 1)^{2} = {(3 \sqrt{18})}^{2} = 9 * 18 = 162$ .

Substitute $y = - x + 3 : (x - 4)^{2} + (- x + 3 + 1)^{2} = 162 \Rightarrow (x - 4)^{2} = 81 => x - 4 = \pm 9$ .

So, $x = 13$ or $x = - 5$ . Then $y = - x + 3$ : If $x = 13, y = - 10$ . If $x = - 5, y = 8.$

Now, since the string is stretched beyond $B$ , the point $B^{'}$ lies in the same direction from $A$ as $B$ .

From $A (4, - 1)$ to $B (1, 2)$ , $x$ decreases and $y$ increases, so we choose $x = - 5, y = 8$ .

On $x$ -axis, $y = 0 \Rightarrow x = 2$ . So $A = (2, 0)$ . Given $B = (4, 2)$ .

Slope of $A B$ is $m = \frac{2 - 0}{4 - 2} = 1$ . So line $A B$ is $y = x - 2$

The line is rotated anticlockwise by $45^{\circ}$ about $A$ .

Angle between original line and new line is $45^{\circ}$ . So the new angle will be $90^{\circ}$ i.e. line is parallel to $y$ and passes through $A$ so new line is parallel to $y$ -axis i.e. $x = 2$ .

After rotation, $B^{'}$ lies on $x = 2$ .

Distance $A B = \sqrt{(4 - 2)^{2} + (2 - 0)^{2}} = \sqrt{8} = 2 \sqrt{2}$ .

So: $(x - 2)^{2} + (y - 0)^{2} = 8$ . Since $x = 2 : y^{2} = 8 \Rightarrow y = \pm 2 \sqrt{2}$

Since rotation is anticlockwise from slope $1$ , the point moves upward from $A$ . So $y > 0$ :

$∴ y = 2 \sqrt{2} \Rightarrow B^{'} = (2, 2 \sqrt{2})$ .

Put $x = - 1$ in floor equation: $- 1 + 2 y = 3 \Rightarrow y = 2$ . So impact point is $P (- 1, 2)$ .

The floor is a straight line, so we use the property: Angle of incidence = angle of reflection.

Incoming path is vertical, so it makes an angle of $90^{\circ}$ with the x-axis.

Now find slope of floor: $x + 2 y = 3 \Rightarrow y = \frac{3 - x}{2}$ . So slope of floor is $- \frac{1}{2}$ .

A line perpendicular to floor has slope $2$ .

Since incidence is vertical, we consider how a vertical direction reflects across a line of slope $- \frac{1}{2}$ .

The reflected direction must satisfy symmetry about the floor, so we construct it geometrically using slope relation:

If one direction is vertical, the reflected direction must make equal angle with the floor on the other side. This gives the new slope:

$m = \frac{3}{4}$ . So rebound path passes through $P (- 1, 2)$ with slope $\frac{3}{4}$ .

Equation of rebound path: $y - 2 = \frac{3}{4} (x + 1)$

Height fallen $= 2 - 1 = 1$

Rebound height $= \frac{2}{3}$ So maximum $y$ after rebound: $y = 2 + \frac{2}{3} = 8 / 3$

Substitute into line: $\frac{8}{3} - 2 = \frac{3}{4} (x + 1) \Rightarrow \frac{2}{3} = \frac{3}{4} (x + 1) \Rightarrow x + 1 = \frac{8}{9} \Rightarrow x = - \frac{1}{9}$

Since motion is constrained by slanted floor $x + 2 y = 3$ , the actual highest point must also satisfy proportional displacement along the reflected line segment above the floor.

Scaling the displacement from $P (- 1, 2)$ in ratio consistent with the $2 : 3$ rebound rule along the oblique direction gives:

$x = - \frac{13}{15}, y = \frac{19}{15}$ .

Line parallel to $3 x - 4 y + 1 = 0$ through $A (4, - 1)$ is: $3 x - 4 y + c = 0$

Substitute $A (4, - 1) : 12 + 4 + c = 0 \Rightarrow c = - 16$

So line is $3 x - 4 y - 16 = 0$ .

Let a point on it be $(x, y)$ and distance from $A (4, - 1)$ be $5 : (x - 4)^{2} + (y + 1)^{2} = 25$

From line: $x = \frac{4 y + 16}{3}$ . Substitute: ${(\frac{4 y + 16}{3} - 4)}^{2} + (y + 1)^{2} = 25$

${(\frac{4 y + 16 - 12}{3})}^{2} + (y + 1)^{2} = 25$ . So $y = 2$ or $y = - 4$

For $y = 2 : x = \frac{8 + 16}{3} = 8$

For $y = - 4 : x = \frac{- 16 + 16}{3} = 0$

We measure distance from $P (3, 5)$ to the line $2 x + 3 y = 14$ along a direction parallel to $x - 2 y = 1$ .

So we move from $(3, 5)$ along a line parallel to $x - 2 y = 1$ until we meet $2 x + 3 y = 14$ .

A line parallel to $x - 2 y = 1$ has form: $x - 2 y = k$

Through $P (3, 5) : 3 - 2.5 = k \Rightarrow k = - 7$

So required line through $P$ is: $x - 2 y = - 7$

Now find intersection with $2 x + 3 y = 14$ .

From $x - 2 y = - 7 \Rightarrow x = 2 y - 7$

Substitute: $2 (2 y - 7) + 3 y = 14 => y = 4$

Then: $x = 2.4 - 7 = 1$

So intersection point is $Q (1, 4)$ .

Now distance $P Q = \sqrt{(1 - 3)^{2} + (4 - 5)^{2}} = \sqrt{5}$ .

We measure the distance from $P (2, 5)$ to the line $3 x + y + 4 = 0$ along a direction parallel to $3 x - 4 y + 8 = 0$ .

A line parallel to $3 x - 4 y + 8 = 0$ is: $3 x - 4 y = k$

Through $P (2, 5) : 3 (2) - 4 (5) = k \Rightarrow 6 - 20 = - 14$ . So $k = - 14$ .

Hence required line through $P$ is: $3 x - 4 y = - 14$

Now find its intersection with $3 x + y + 4 = 0$ .

From $3 x - 4 y = - 14 : 3 x = 4 y - 14 \Rightarrow x = \frac{4 y - 14}{2}$

Substitute into $3 x + y + 4 = 0 : 3 (\frac{4 y - 14}{3}) + y + 4 = 0$

$(4 y - 14) + y + 4 = 0 \Rightarrow 5 y - 10 = 0 \Rightarrow y = 2$

Then: $x = \frac{8 - 14}{3} = - 2$ . So intersection point is $Q (- 2, 2)$ .

Now distance $P Q = \sqrt{(- 2 - 2)^{2} + (2 - 5)^{2}} = 5$ .

Let $A (1, 3)$ and $C (5, 1)$ be opposite vertices of a rectangle.

Midpoint of diagonal: $M = (\frac{1 + 5}{2}, \frac{3 + 1}{2}) = (3, 2)$ . So centre is $(3, 2)$ .

Let other vertices be $B$ and $D$ on line $y = 2 x + c$ .

Since diagonals bisect each other, $B$ and $D$ are symmetric about $(3, 2)$ .

So if $B (x, y)$ is on the line, then $D (6 - x, 4 - y)$ is also on it.

For $B : y = 2 x + c$ . For $D : 4 - y = 2 (6 - x) + c$

Substitute $y : 4 - (2 x + c) = 12 - 2 x + c \Rightarrow c = - 4$

So line is $y = 2 x - 4$ . Let $x = 2$ , then $y = 0$ so $B (2, 0)$ .

The other two vertices must lie on this line $y = 2 x - 4$ and must be symmetric about the midpoint $M (3, 2)$ .

So if we pick any point $B (x, y)$ on the line, its opposite vertex is automatically fixed by midpoint symmetry: $D = (6 - x, 4 - y)$

Now both $B$ and $D$ will always satisfy the line equation, so we are free to choose any value of $x$ that makes calculations simple.

We chose $x = 2$ because it avoids fractions and gives: $y = 2.2 - 4 = 0$

So $B = (2, 0)$ is an easy clean point on the line, and then: $D = (6 - 2, 4 - 0) = (4, 4)$ .

Thus, $c = - 4$ .

Let the line through $(x^{'}, y^{'})$ make an angle $α$ with the $x$ -axis. Its parametric form is:

$x = x^{'} + r \cos α, y = y^{'} + r \sin α$ , where $r$ is the distance measured along the line.

Substitute into $A x + B y + C = 0$ :

$A (x^{'} + r \cos α) + B (y^{'} + r \sin α) + C = 0, A x^{'} + B y^{'} + C + r (A \cos α + B \sin α) = 0$

Solve for $r : r (A \cos α + B \sin α) = - (A x^{'} + B y^{'} + C)$

$r = - (A x^{'} + B y^{'} + C) / (A \cos α + B \sin α)$

Since length is absolute value of displacement: Length $= | r |$

Length $= | \frac{A x^{'} + B y^{'} + C}{A \cos α + B \sin α} |$ .

Let the required line through $P (1, 2)$ be $y - 2 = m (x - 1)$

Parametric form: $x = 1 + t, y = 2 + m t$ . So at $P, t = 0$ .

Intersection with $x + y - 5 = 0 : (1 + t) + (2 + m t) - 5 = 0$

$(m + 1) t - 2 = 0 => t_{A} = \frac{2}{m + 1}$

Intersection with $2 x - y = 7 : 2 (1 + t) - (2 + m t) = 7 \Rightarrow t_{B} = \frac{7}{2 - m}$

Since A and B lie on the same side of $P : t_{A}, t_{B} > 0 \Rightarrow - 1 < m < 2$

Scale factor along the line is $\sqrt{1 + m^{2}}$

$P A = \sqrt{1 + m^{2}} t_{A}, P B = \sqrt{1 + m^{2}} t_{B}$

Harmonic mean condition: $= \frac{2}{\frac{1}{P A} + \frac{1}{P B}} = 10$

$= \frac{2}{\frac{1}{s t_{A}} + \frac{1}{s t_{B}}} = 10$ , where $s = \sqrt{1 + m^{2}}$

$= \frac{2 a}{\frac{1}{t_{1}} + \frac{1}{t_{B}}} = 10$

$\frac{1}{t_{A}} = \frac{m + 1}{2}, \frac{1}{t_{B}} = \frac{2 - m}{7}$

Sum: $\frac{1}{t_{A}} + \frac{1}{t_{B}} = \frac{5 m + 11}{14}$

$\Rightarrow \frac{2 s}{\frac{5 m + 11}{14}} = 10$

$\frac{14 x}{5 m + 11} = 5$ . Substitute $s = \sqrt{1 + m^{2}}$ :

$14 \sqrt{1 + m^{2}} = 25 m + 55 \Rightarrow 196 (1 + m^{2}) = (25 m + 55)^{2}$

Solve: $m = \frac{- 2750 \pm \sqrt{2707936}}{858}$

Required line: $y - 2 = m (x - 1)$ .

Any point on the line is $(- 2 + r \cos θ, - 3 + r \sin θ)$ . Let $B = (- 2 + r_{1} \cos θ, - 3 + r_{1} \sin θ)$ and $C = (- 2 + r_{2} \cos θ, - 3 + r_{2} \sin θ)$ .

Then $A B = | r_{1} |$ and $A C = | r_{2} |$

$∴ - 2 + r_{1} \cos θ + 3 (- 3 + r \sin θ) = 9 \Rightarrow r_{1} = (20) (c o s θ + 3 \sin θ)$

and $r_{2} = \frac{4}{\cos θ + \sin θ}$

Given that $A B . A C = 20 \Rightarrow r_{1} . r_{2} = \pm 20 \Rightarrow \cos^{2} θ + 3 \sin^{2} θ + 4 \sin θ \cos θ = 4$

$=> \tan θ = 3, 1$ . So the possible lines are $x - y = 1, 3 x - y + 3 = 0$ .

Let the line through $P (3, 4)$ making angle $θ$ with the positive $x$ -axis be

$(x, y) = (3, 4) + r (\cos θ, \sin θ)$

Substitute into $y^{2} = 4 x : (4 + r \sin θ)^{2} = 4 (3 + r \cos θ)$

$\Rightarrow 16 + 8 r \sin θ + r^{2} \sin^{2} θ = 12 + 4 r \cos θ \Rightarrow r^{2} \sin^{2} θ + 8 r \sin θ - 4 r \cos θ + 4 = 0$

$\Rightarrow r^{2} \sin^{2} θ + 4 r (2 \sin θ - \cos θ) + 4 = 0$ .

The midpoint of $B C$ is $M = (\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2})$ .

So, the equation of the median from $A$ is the line passing through $A$ and $M$ :

$| \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ \frac{x_{2} + x_{3}}{2} & \frac{y_{2} + y_{3}}{1} & 1 \end{matrix} | = 0$

Using linearity of determinants:

$| \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{matrix} | + | \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} | = 0$ .

Slope of $x - 2 y + 3 = 0$ is $m_{1} = \frac{- 1}{- 2}) = \frac{1}{2}$ and slope of $3 x + y - 1 = 0$ is $- 3$ .

If $θ$ is the acute angle between them then $\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | = 7$ .

So the acute angle between them is $\tan^{- 1} 7$ and obtuse angle is $π - \tan^{- 1} 7$ .

Slope of $x + y = 3$ is $m_{1} = - 1$ . Slope of the line passing through $(1, 1)$ and $(- 3, 4)$ is $m_{2} = - \frac{3}{4}$ .

If $θ$ is the acute angle between them then $\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | = \frac{1}{7}$ .

So the acute angle between them is $\tan^{- 1} \frac{1}{7}$ and obtuse angle is $π - \tan^{- 1} \frac{1}{7}$ .

Slope of $2 x + 3 y + 4 + k (6 x - y + 12) = 0$ is $m_{1} = - \frac{2 + 8 k}{3 - k} = \frac{2 + 6 k}{k - 3}$ and slope of the line $7 x + 5 y - 4 = 0$ is $m_{2} = - \frac{7}{5}$ .

Since the lines are perpendicular so $m_{1} m_{2} = - 1 \Rightarrow k = - \frac{29}{37}$ .

Let $A (x_{1}, y_{1}), B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ be the three vertices of a $△ A B C$ . Let $P$ and $Q$ represent the mid-points of the sides $A B$ and $A C$ respectively.

$P = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ and $Q = (\frac{x_{1} + x_{3}}{2}, \frac{y_{1} + y_{3}}{2})$

Slope of line $P Q : m_{1} = \frac{\frac{y_{1} + y_{3}}{2} - \frac{y_{1} + y_{2}}{2}}{\frac{x_{1} + x_{3}}{2} - \frac{x_{1} + x_{2}}{2}} = \frac{y_{3} - y_{2}}{x_{3} - x_{2}}$

Slope of line $B C : m_{2} = \frac{y_{3} - y_{2}}{x_{3} - x_{2}}$ .

Slope of $A B : m_{1} = \frac{0 - 2}{2 - 0} = - 1$ . Slope of $C D, m_{2} = \frac{y - 7}{x}$ .

Since $A B ∥ C D$ , therefore, $m_{1} = m_{2} \Rightarrow y = 7 - x$ .

Since the trapezium is isosceles, therefore, $A D = B C \Rightarrow (x - 2)^{2} + y^{2} = 5 \Rightarrow x = 7, 2 \Rightarrow y = 0, 5$ .

Slope of the given lines are $m_{1} = - \frac{a + b}{a - b}, m_{2} = - \frac{a - b}{a + b}$ and $m_{3} = - 1$ respectively.

If angle between first and third line is $α$ then $\tan α = \frac{b}{a}$ .

If angle between second and third line is $β$ then $\tan β = \frac{b}{a}$ .

Since both the angles are same we have an isosceles triangle. Let $θ$ be the vertical angle.

Then $θ + α + β = 180^{\circ} \Rightarrow θ = 180^{\circ} - 2 α \Rightarrow \frac{θ}{2} = 90^{\circ} - α$

$\tan \frac{θ}{2} = 2 \cot α \Rightarrow θ = 2 \tan^{- 1} \frac{a}{b}$ .

Slope of $x = a$ is $\tan 90^{\circ}$ and slope of the line is $b y + c = 0$ is $\tan 0^{\circ}$ . Thus, angle between the two lines is $90^{\circ}$ .

Equation of the line whose intercepts are $3, 4$ is given by $\frac{x}{3} + \frac{y}{4} = 1$ and has a slope of $m_{1} = - \frac{4}{3}$ .

Equation of the line whose intercepts are $1, 8$ is given by $\frac{x}{1} + \frac{y}{8} = 1$ and has a slope of $m_{2} = - 8$

The angle is given by $θ$ then $\tan θ = | \frac{- \frac{4}{3} + 8}{1 + \frac{32}{3}} | = \frac{4}{7} \Rightarrow θ = \tan^{- 1} \frac{4}{7}$ .

Slope of the line $\frac{x}{a} + \frac{y}{b} = 1$ is $m_{1} = - \frac{b}{a}$ and slop of the line $\frac{x}{b} - \frac{y}{a} = 1$ is $m_{2} = \frac{a}{b}$ .

Clearly, $m_{1} m_{2} = - 1$ , and thus, the two lines are perpendicular to each other.

Slope of the line joining $(2, - 3)$ and $(- 1, 2)$ is $m_{1} = \frac{2 + 3}{- 1 - 2} = - \frac{5}{3}$ .

Slope of the line joining $(3, 7)$ and $(- 2, 4)$ is $m_{2} = \frac{4 - 7}{- 2 - 3} = \frac{2}{5}$ .

Clearly, $m_{1} m_{2} = - 1$ , and thus, the two lines are perpendicular to each other.

Slope of the line joining $(a, 2 a)$ and $(- 2, 3)$ is $m_{1} = \frac{3 - 2 a}{- 2 - a}$ .

Slope of the line $4 x + 3 y + 5 = 0$ is $m_{2} = - \frac{4}{3}$ .

Given that lines are perpendicular to each other therefore $m_{1} m_{2} = - 1$

$\Rightarrow \frac{3 - 2 a}{- 2 - a} . \frac{- 4}{3} = - 1 \Rightarrow 12 - 8 a = - 6 - 3 a \Rightarrow a = \frac{15}{8}$ .

Both the lines have the same slope of $7$ , thus lines are parallel to each other.

Slope of the line $k^{2} x + k y + 1 = 0$ is $- \frac{k^{2}}{k} = - k$ .

Slope of the line $x - k y = 1$ is $\frac{1}{k}$ . Clearly, product of the slopes is $- 1$ i.e. lines are perpendicular to each other.

Slope of the line $x - y + 2 + k (2 x + 3 y) = 0$ is $- \frac{2 k + 1}{3 k - 1} \Rightarrow \frac{2 k + 1}{1 - 3 k}$ .

Slope of the line $3 x + y = 0$ is $- 3$ .

Because the lines are parallel the slopes will be equal. $\Rightarrow 2 k + 1 = 9 k - 3 \Rightarrow k = \frac{4}{7}$ .

First and third lines have same slope, and, second and fourth lines have same slope. Thus, they will form a parallelogram.

Slope of the line $x \cos θ + y \sin θ = 2$ is $m_{1} = - \frac{\cos θ}{\sin θ} = - \cot θ$ .

Slope of the line $x - y = 3$ is $m_{2} = 1$ .

Since the lines are perpendicular $m_{1} m_{2} = - 1 \Rightarrow \cot θ = 1 \Rightarrow θ = 45^{\circ}$ .

Slope of the line $x - 3 y + 5 + k (x + y - 3) = 0$ is $m_{1} = \frac{k + 1}{3 - k}$ . Slope of the line $x + y = 1$ is $- 1$ .

Since the lines are perpendicular $\Rightarrow m_{1} m_{2} = - 1 \Rightarrow k + 1 = 3 - k \Rightarrow k = 1$ .

Thus, equation of the first line becomes $2 x - 2 y + 2 = 0 \Rightarrow x - y + 1 = 0$ .

Let $A = (0, 0), B = (a, 0), C = (\frac{a}{2}, \frac{\sqrt{3} a}{2})$ . (You can get these points by rotating the line moving by a distance $a$ along that line)

Midpoint of $A B : M = (\frac{0 + a}{2}, \frac{0 + 0}{2}) = (\frac{a}{2}, 0)$

Slope of $A B : m_{A B} = \frac{0 - 0}{a - 0} = 0$

Since $x_{C} = x_{M} = \frac{a}{2}$ , the line $C M$ is vertical.

A vertical line is perpendicular to a horizontal line $A B$ . Hence $C M ⊥ A B$ .

Place the rhombus with vertices $A = (a, 0), B = (0, b), C = (- a, 0), D = (0, - b)$ for $a, b > 0$ .

Each side has length $| A B | = \sqrt{(a - 0)^{2} + (0 - b)^{2}} = \sqrt{a^{2} + b^{2}}$ , and by symmetry all four sides are equal, confirming $A B C D$ is a rhombus.

The diagonal $A C$ runs from $(a, 0)$ to $(- a, 0)$ , so its slope is $m_{A C} = \frac{0 - 0}{- a - a} = 0$ .

The diagonal $B D$ runs from $(0, b)$ to $(0, - b)$ , so it is a vertical line with undefined slope, meaning it is parallel to the $y$ -axis.

A line with slope $0$ is horizontal, and a vertical line is perpendicular to every horizontal line. Therefore $A C ⊥ B D$ .

Equation of the line parallel to the given line is $3 x - y + k = 0$ . Also, given that this passes through $(3, 4)$ ; putting the point in the equation

$3.3 - 4 + k = 0 \Rightarrow k = - 5$ .

Thus, equation of the required line is $3 x - y - 5 = 0$ .

The line perpendicular to the line $4 x - 3 y = 10$ is given by $3 x + 4 y + k = 0$ .

Given that it passes through $(2, 3)$ ; putting this point in the equation

$2.3 + 4.3 + k = 0 \Rightarrow k = - 18$ .

Thus, equation of the required line is $3 x + 4 y - 18 = 0$ .

Since the intercept is $\frac{4}{3}$ on $y$ -axis, therefore, the line passes through $(0, \frac{4}{3})$ . Also lines perpendicular to the line $3 - 4 y + 11 = 0$ is given by

$4 x + 3 y + k = 0$ . Putting $(0, \frac{4}{3})$ gives us $k = - 4$ .

Thus, required line is $4 x + 3 y - 4 = 0$ .

Mid-point of $(1, 1)$ and $(2, 3)$ is $(\frac{3}{2}, 2)$ . Equation of the line passing though these two points is

$y - 1 = \frac{3 - 1}{2 - 1} (x - 1) \Rightarrow y - 1 = 2 x - 2 \Rightarrow 2 x - y = 0$ .

Line perpendicular to this wll be $x + 2 y + k = 0$ , which passes through the mid-point. Thus,

$\frac{3}{2} + 4 + k = 0 \Rightarrow k = - \frac{11}{2}$ .

Thus, equation of the line is $2 x + 4 y - 11 = 0$ .

Given line is $x + y = a$ , which cuts the $x$ and $y$ axes at $A$ and $B$ respectively. $\Rightarrow A = (a, 0)$ and $B = (0, a)$ .

Let $\frac{A N}{N K} = k$ , then $N = (\frac{a}{1 + k}, \frac{k a}{1 + k})$ .

Since line $M N$ is perpendicular to $A B$ and it passes through $N$ , therefore,

$x - y - (\frac{a}{1 + k} - \frac{k a}{1 + k}) = 0 \Rightarrow x - y = \frac{1 - k}{1 + k} a$

This line cuts the $y$ -axis at $M$ . THus, $M = (0, \frac{k - 1}{k + 1} a)$

$Δ A M N = \frac{1}{2} . A N . N M$

$A N^{2} = {(a - \frac{a}{k + 1})}^{2} + {(\frac{0 - k a}{1 + k})}^{2} \Rightarrow A N = \frac{\sqrt{2} k a}{1 + k}$

$N M = \frac{\sqrt{2} a}{1 + k}$ .

Thus, $Δ A M N = \frac{a^{2} k}{(1 + k^{2})}$ and $Δ O A B = \frac{1}{2} . a^{2}$

Thus, $Δ A M N : Δ O A B = 3 : 8 \Rightarrow 3 (1 + k)^{2} = 16 k \Rightarrow k = 3, \frac{1}{3}$ but if $k = \frac{1}{3}$ then $M$ lies outside of $O B$ . Thus, $k = 3$ .

Figure 9.10.

Slope of the line $3 x - y + 5 = 0$ is $3$ . Let the slope of the required line is $m$ then

$\tan 45^{\circ} = | \frac{m - 3}{1 + 3 m} | \Rightarrow (m - 3) / (1 + m) = \pm 1 ∴ m = - 2, \frac{1}{2}$ .

Slope of the given line is $\frac{1}{2}$ . Let $m$ be the slope of the line passing through $(3, 2)$ and making an angle of $45^{\circ}$ with the line.

$\Rightarrow \tan 45^{\circ} = | \frac{\frac{1}{2} - m}{1 + \frac{m}{2}} | \Rightarrow m = 3, - \frac{1}{2}$ .

Thus, equations of required line are $y - 2 = 3 (x - 3)$ and $y - 2 = - \frac{1}{2} (x - 3)$ i.e. $3 x - y = 7$ and $x + 3 y = 9$ .

Given line is $x + y - 2 = 0$ , its slope, $m_{1} = - 1$ .

Let the slope of the line which makes an angle of $60^{\circ}$ with this line be $m_{1}$ , then

$\tan 60^{\circ} = | \frac{m_{1} - n}{1 + m . m_{1}} | \Rightarrow \sqrt{3} = | \frac{- 1 - m}{1 - m} |$

$m = 2 + \sqrt{3}, 2 - \sqrt{3}$ .

Thus, equation of two other sides of the triangle are

$y - 3 = (2 + \sqrt{3}) (x - 2)$ and $y - 3 = (2 - \sqrt{3}) (x - 2)$ .

Given that equation of $B C$ is $3 x - 4 y + 1 = 0$ and the equation of $A B$ is $4 x + y - 1 = 0$ .

Since $A B = A C$ , therefore, $∠ A B C = ∠ A C B = α$ (say)

Slope of the line $B C = \frac{3}{4}$ and slope of $A B = - 4$ . Let slope of $A C = m$ .

Thus, $\frac{- 4 - \frac{3}{4}}{1 - \frac{4.3}{4}} = \frac{\frac{3}{4} - m}{1 + \frac{2}{4} m} \Rightarrow m = - \frac{52}{89}$

Thus, equation of $A C$ is $y + 7 = - \frac{52}{89} (x - 2) \Rightarrow 52 x + 89 y + 519 = 0$ .

Figure 9.11.

Equation of line through $(- 2, - 7)$ is given by $y + 7 = m (x + 2) \Rightarrow y = m x + 2 m - 7$ . This line cuts the given lines at $A$ and $B$ respectively. Solving the equations gives

$A = (\frac{33 - 6 m}{4 + 3 m}, \frac{20 m - 28}{4 + 3 m})$ and $B = (\frac{24 - 4 m}{4 + 3 m}, \frac{11 m - 28}{4 + 3 m})$

According to question $A B = 3 \Rightarrow A B^{2} = 9 \Rightarrow \frac{81}{(4 + 3 m)^{2}} + \frac{81 m^{2}}{(4 + 3 m)^{2}} = 9$ .

$\Rightarrow 9 + 9 m^{2} = 16 + 9 m^{2} + 24 m \Rightarrow m = - \frac{7}{24}$ .

When $m \to \infty$ both sides become tend to $9$ i.e. line may be perpendicular to $x$ -axis.

Thus, equations of the required lines are $x = - 2$ and $y + 7 = - \frac{7}{24} (x + 2) \Rightarrow 7 x + 24 y + 182 = 0$ .

The line parallel to $x + 2 y = 3$ is given by $x + 2 y = k$ . Since it passes through $(3, 4)$ , therefore,

$k = 3 + 2.4 = 11$ . Hence, equation of the required line becomes $x + 2 y = 11$ .

The line parallel to $3 x + 4 y = 12$ is given by $3 x + 4 y = k$ . Since it passes through $(4, 3)$ , therefore,

$k = 3.4 + 4.3 = 24$ . Hence, equation of the required line becomes $3 x + 4 y = 24$ .

Equation of the straight line parallel to $3 x - 4 y + 6 = 0$ is given by $3 x - 4 y + k = 0$ . It passes through the mid-point of the line segment made by $(2, 3)$ and $(4, - 1)$ i.e. $(3, 1)$ .

Thus, $3.3 - 4.1 + k = 0 \Rightarrow k = - 5$ . Hence, the equation of the required line is $3 x - 4 y - 5 = 0$ .

Euation of the line joining the points $(2, 3)$ and $(3, - 1)$ is given by

$y - 3 = \frac{- 1 - 3}{3 - 2} (x - 2) \Rightarrow y - 3 = 8 - 4 x \Rightarrow 4 x + y = 11$

A line parallel to above line wil be $4 x + y = k$ . Since it passes through $(2, 1)$ , therefore,

$4.2 + 1 = k \Rightarrow k = 9$ . So the required line becomes $4 x + y = 9$ .

Equation of the line parallel to the line $l x + m y + n = 0$ is given by $l x + m y + k = 0$ .

Since it passes through $(α, β)$ , therefore, $l α + m β + k = 0$ .

Thus, equation of required line is $l x + m y - (l α + m β) = 0$ .

Equation of the line perpendicular to the line $2 x + 5 y = 31$ is $5 x - 2 y + k = 0$ . Since it passes through $(2, 5)$ , therefore,

$5.2 - 2.5 + k = 0 \Rightarrow k = 0$ . So the required line is $5 x - 2 y = 0$ .

Any line perpendicular to the given line is $2 a y + x y^{'} + k = 0$ . Since it passing through $(x^{'}, y^{'})$ , therefore,

$k = - 2 a y^{'} - x^{'} y^{'}$ . Thus, required line is $2 a (y - y^{'}) + y^{'} (x - x^{'}) = 0$ .

Slope of the first line is x $m_{1} = \frac{m n + n^{2}}{m^{2} - m n}$ , and the slope of the second line is $m_{2} = \frac{m n - n^{2}}{m n + m^{2}}$ .

Let $θ$ be the angle between these lines then $\tan θ = | \frac{\frac{m n + n^{2}}{m^{2} - m n} - \frac{m n - n^{2}}{m n + m^{2}}}{1 + \frac{m n + n^{2}}{m^{2} - m n} . \frac{m n - n^{2}}{m n + m^{2}}} |$

$\Rightarrow \tan θ = \frac{4 m^{2} n^{2}}{m^{4} - n^{4}} \Rightarrow θ = \tan^{- 1} \frac{4 m^{2} n^{2}}{m^{4} - n^{4}}$ .

Any line perpendicular to the line $x \sec θ + y \csc θ = a$ is given by $x \csc θ - y \sec θ = k$ , however, the line passes through $(a \cos^{3} θ, a \sin^{3} θ)$ , therefore,

$a \cos^{3} θ \csc θ - a \sin^{3} θ \sec θ = k = \frac{a \cos^{4} θ - a \sin^{4} θ}{\sin θ \cos θ} = \frac{a \cos 2 θ}{\sin θ \cos θ}$

Thus, the given line becomes $x \cos θ - y \sin θ = a \cos 2 θ$ .

Any line perpendicular to the line $\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$ is given by $\frac{x}{b} \sin θ - \frac{y}{a} \cos θ = k$ . Since this new line passes through $(a \cos θ, b \sin θ)$ , therefore,

$k = \frac{a \cos θ . \sin θ}{b} - \frac{b \sin θ . \cos θ}{a}$ . Thus, equation of the new line becomes

$a x \sec θ - b y \csc θ = a^{2} - b^{2}$ .

Let the parallelogram be $A B C D$ .

Figure 9.12.

Let the equations of sides $A B$ and $A D$ of the parallelogram be $4 x + 5 y = 0$ and $7 x + 2 y = 0$ . Solving these equations gives $A = (0, 0)$ .

Equation of one of the diagonals of the parallelogram is $11 x + 7 y = 9$ , which does not pass through $A$ so it must be the diagonal $B D$ .

Solving $A D$ and $B D$ and $A B$ and $B D$ gives us $B = (5 / 3, - 4 / 3)$ and $D = (- 2 / 3, 7 / 3)$ .

Thus, mid-point of diagonals is $H = (\frac{1}{2}, \frac{1}{2})$ . Thus, equation of the other diagonal which passes through $A$ and $H$ is $x = y$ .

Solving the three equations pairwise gives us three coordinates
$A = (\frac{c - c_{1}}{m_{1} - m_{2}}, \frac{m_{1} c_{2} - m_{2} c_{1}}{m_{1} - m_{2}}), B = (0, c_{1})$ and $C = (0, c_{2})$ .

Putting these points in the formula for area of triangle gives us
$| \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{1})] |$

$= \frac{1}{2} \frac{c_{2} - c_{1}}{m_{1} - m_{2}} (c_{1} - c_{2}) = \frac{1}{2} . (c_{2} - c_{1})^{2} / (m_{1} . m_{2})$ .

Let the three lines be $B C, C A$ and $A B$ whose equations are $y = m_{1} x + c_{1}, y = m_{2} x + c_{2}$ and $y = m_{3} x + c_{3}$ .

Let the lines $B C, C A$ and $A B$ meet $y$ -axis at $P, Q$ and $R$ respectively. From figure

$Δ A B C = Δ A Q R - Δ B P R + Δ C P Q$

Proceeding like previous problem we have the required result.

Figure 9.13.

Let $A (x_{1}, y_{1})$ be the point of intersection of first two equations, $B (x_{2}, y_{2})$ that of second and third, and $C (x_{3}, y_{3})$ that of first and last equations.

$Δ_{A B C} = \frac{1}{2} | \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} | = \frac{1}{2} \frac{| \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} | \times | \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} |}{| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} |}$

$= \frac{1}{2} | \begin{matrix} a_{1} x_{1} + b_{1} y_{1} + c_{1} & a_{2} x_{1} + b_{2} y_{1} + c_{2} & a_{3} x_{1} + b_{3} y_{1} + c_{3} \\ a_{1} x_{2} + b_{1} y_{2} + c_{1} & a_{2} x_{2} + b_{2} y_{2} + c_{2} & a_{3} x_{2} + b_{3} y_{2} + c_{2} \\ a_{1} x_{3} + b_{1} y_{3} + c_{1} & a_{2} x_{3} + b_{2} y_{3} + c_{+} 2 & a_{3} x_{3} + b_{3} y_{3} + c_{3} \end{matrix} | \div Δ$

$= \frac{1}{2} | \begin{matrix} 0 & 0 & a_{3} x_{1} + b_{3} y_{1} + c_{3} \\ a_{1} x_{2} + b_{2} y_{2} + c_{1} & 0 & 0 \\ 0 & a_{2} x_{3} + b_{2} y_{3} + c_{2} & 0 \end{matrix} | \div Δ$

$(x_{1}, y_{1})$ satisfied the above equation and also $a_{1} x_{1} + b_{2} y_{1} + c_{1} = 0$ and $a_{2} x_{1} + b_{2} y_{1} + c_{2} = 0$

Let $a_{3} x_{1} + b_{3} y_{1} + c_{3} = λ_{1} \Rightarrow a_{3} x_{1} + b_{3} y_{1} + c_{3} - λ = 0$

Thus, eliminating $(x_{1}, y_{1})$ from these equations gives us

$| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} - λ \end{matrix} | = 0 \Rightarrow Δ - λ | \begin{matrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{matrix} |$

$\Rightarrow Δ - λ_{1} C_{3} = 0 \Rightarrow λ_{1} = Δ / C_{3}$ , where $C_{3}$ is cofactor of $c_{1}$ . Similarly, $λ_{2} = Δ / C_{1}$ and $λ_{3} = Δ / C_{2}$

Thus, area of the required triangle is $| \frac{1}{2} (λ_{1} λ_{2} λ_{3} \div Δ) | = \frac{Δ^{2}}{2 | C_{1} C_{2} C_{3} |}$

The equation of any line with gradient $2$ is given by $y = 2 x + c$ . This line will intersect with given lines at $A = ((\frac{3}{2} - \frac{x}{6}, 3 = \frac{2 c}{3}), B = (1 - \frac{2 c}{3}, 2 - \frac{c}{3})$ and $C = (2 + \frac{c}{3}, 4 + \frac{5 c}{3})$ .

Clearly, $A$ is the middle point of $B C$ . Thus, intercepts are equal.

Let $(p, q)$ be the foot of the perpendicular. Then $q = 3 p + 4$ . Also, line perpendicular to it will have the equation $x + 3 y = k$ .

Since $x + 3 y = k$ will pass through $(2, 3)$ , therefore, $k = 11$ . Also, $(p, q)$ will lie on this line so $p + 3 q = 11$ .

Solving the two equations gives us $(p, q) = (- \frac{1}{10}, \frac{37}{10})$ .

Equation of the line mirror $A B$ is $4 x + 7 y + 13 = 0$ . Let $P = (- 8, 12)$ and $Q = (a l p h a, b e t a)$ be the image of $P$ in the line mirror $A B$ . Then $P Q ⊥ A B$ and $P L = L Q$ .

Thus equation of $P L$ would be $7 x - 4 y = k$ . Since it passes through $P (- 8, 12)$ , therefore,

$k = - 56 - 48 = - 104$ . So the equation of $P L$ is $7 x - 4 y + 104 = 0$ . Solving the equations $L = (- 12, 5) . k = - 56 - 48 = - 104$ . So the equation of $P L$ is $7 x - 4 y + 104 = 0$ . Solving the equations $L = (- 12, 5)$ .

Since $Q$ is mirror image of $P$ so $L$ will be midpoint of $P Q$ . Thus, $Q = (- 16, - 2)$ .

Figure 9.14.

Slope of $A B = - \frac{a}{b}$ and that of $P Q = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

Since $A B ⊥ P Q \Rightarrow - \frac{a}{b} . \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = - 1 \Rightarrow \frac{x_{2} - x_{1}}{a} = \frac{y_{2} - y_{1}}{b} = k$ (say)

$\Rightarrow x_{2} - x_{1} = k a, y_{2} - y_{1} = k b \Rightarrow a (x_{2} - x_{1}) + b (y_{2} - y_{1}) = k (a^{2} + b^{2})$

Like previous problem $L$ is the midpoint of $P Q$ i.e. $L = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ and it also lies on the line $A B$ , therefore

$a \frac{x_{1} + x_{2}}{2} + b \frac{y_{1} + y_{2}}{2} + c = 0 \Rightarrow a x_{2} + b y_{2} + c = - (a x_{1} + b y_{1} + c)$

From previously obtained equation $- 2 (a x_{1} + b y_{1} + c) = k (a^{2} + b^{2})$

Thus, $\frac{x_{2} - x_{1}}{a} = \frac{y_{2} - y_{1}}{b} = - \frac{2 (a x_{1} + b y_{1} + c_{1})}{a^{2} + b^{2}}$ .

Figure 9.15.

Given equation of horizontal line is $3 x - 2 y - 5 = 0$ and equation of $P A$ is $x - 2 y - 3 = 0$ .

Solving these two equations gives us $A = (1, - 1)$ .

Figure 9.16.

Let slope of $A Q = m$ . Slope of horizontal line is $\frac{3}{2}$ and slope of $A P = \frac{1}{2}$ .

Let $∠ L A P = α$ then $∠ Q A M = α$ .

$∴ \tan α = 4 / 7, | \frac{2 m - 3}{2 + 3 m} |$

Thus, $m = 1 / 2, (29) / 2$ . So slope of $A Q = \frac{29}{2}$ , and hence, equation of $A Q$ is given by

$y + 1 = \frac{29}{2} (x - 1) \Rightarrow 29 x - 2 y - 31 = 0$ .

Since the light travels through the shorted path $P R$ must be the incident ray and $R Q$ should be the reflected ray. If $S$ be the image of $P$ w.r.t. line mirror $2 x + y = 7$ , then $P R + R Q = S R + R Q$ .

Thus, $P R + R Q$ will be least when $S R + R Q$ will be least i.e. when point $Q, R, S$ are collinear.

Figure 9.17.

Given equation of the line $L M$ is $2 x + y = 7$ . Equation of $P H$ would be $x - 2 y = k$ , which passes through $(- 3, 4)$ so $k = - 3 - 8 = - 11$

Solving the two equations gives $H = (\frac{3}{5}, \frac{29}{5})$ . Let $S = (α, β)$ then

$\frac{3}{5} = \frac{α - 3}{2} \Rightarrow α = \frac{21}{5}$ and $β = \frac{38}{5}$

Equation of $S Q$ is $y - 1 = \frac{\frac{38}{5} - 1}{\frac{21}{5}} x \Rightarrow 11 x - 17 y + 7 = 0$ .

$Q R$ is the refracted ray. According to question, $∠ S Q R = 15^{\circ}$ . From given equations we get $Q = (\frac{8}{5}, - \frac{3}{5})$ .

Figure 9.18.

Slope of $Q P = \frac{2}{3}$ , so slope of $Q S = \frac{2}{3}$ . Slope of $A B = - 1$

Let $∠ P Q B = θ = ∠ A Q S$ , then $\tan θ = | \frac{\frac{2}{3} + 1}{1 - \frac{2}{3}} |$ .

Let slope of $Q R = m$ . $∵ ∠ S Q R = 15^{\circ} ∴ \tan 15^{\circ} = | \frac{\frac{2}{3} - m}{1 + \frac{2}{3} m} | = 2 - \sqrt{3} => m = \frac{3 \sqrt{3} - 4}{7 - 2 \sqrt{3}}, \frac{3 \sqrt{3} - 8}{1 - 2 \sqrt{3}}$ .

Any line passing through the point of intersection of given lines is given by $5 x - y - 9 + k (x + 6 y - 8) = 0$ .

Given that this line passes through $(2, - 2)$ , then $2 + 2 - 9 + k (2 - 12 - 8) = 0 \Rightarrow k = \frac{1}{6}$ .

So by putting $k$ back in the equaiotn to obtain $x - 2 = 0$ .

Equation of the line passing through the point of intersection of given lines is $x - y - 1 + k (2 x - 3 y + 1) = 0$

Slope of this line is $\frac{1 + 2 k}{1 + 3 k}$ . Given that this new line is parallel to $3 x + 4 y - 14 = 0$

Thus, $\frac{1 + 2 k}{1 + 3 k} = - \frac{3}{4} \Rightarrow k = - \frac{7}{17}$ . Putting this value of $k$ in the equation above gives us the required equation as $3 x + 4 y - 24 = 0$ .

Equation of the line passing through the point of intersection of given lines is $3 x - 4 y - 7 + k (12 x - 5 y - 13) = 0$

Slope of this line is $\frac{3 + 12 k}{4 + 5 k}$ . Given that this new line is perpendicular to $2 x - 3 y + 5 = 0$ so the new line's slope must be equal to $\frac{3}{2}$ .

Thus, $\frac{3 + 12 k}{4 + 5 k} = \frac{3}{2} \Rightarrow k = - \frac{6}{12}$ . Putting this value of $k$ in the equation above gives us the required equation as $33 x + 22 y + 13 = 0$ .

Equation of the line passing through the point of intersection of given lines is $x + 3 y + 4 + k (3 x + y + 4) = 0$ .

Slope of this line $- \frac{1 + 3 k}{3 + k}$ . We know that slope of the lines equally inclined to axes are $\pm 1$ .

Equating: $- \frac{1 + 3 k}{3 + k} = \pm 1 \Rightarrow k = - 1, 1$

Thus, required lines are $x - y = 0$ and $x + y + 2 = 0$ .

Let the equation of line $A B$ and $A C$ be $3 x - 2 y + 6 = 0$ and $4 x + 5 y - 20 = 0$ .

Figure 9.19.

Since $B M$ passes through the orthocenter $H (1, 1)$ and is perpendicular to $A C$ , therefore, equation of $B M$ will be $5 x - 4 y - (5.1 - 4.1) = 0 \Rightarrow 5 x - 4 y - 1 = 0$

Similarly equation of $C N$ will be $2 x + 3 y - 5 = 0$

Solving $A B$ and $B M$ gives us $B = (- 13, - \frac{33}{2})$ and solving $A C$ and $C N$ gives $C = (\frac{35}{2}, - 10)$ .

Thus, equation of $B C$ will be $26 x - 122 y - 1675 = 0$ .

Let $A B C D$ be a parallelogram such that $A B$ is $u = p, B C$ is $v = s, C D$ is $u = q$ , and $A D$ is $v = r$ .

Equation of $A C$ , which passes through the point of intersection of lines $u - p = 0$ and $v - r = 0$ is

$u - p + k (v - r) = 0 \Rightarrow a x + b y + c - p + k (a^{'} x + b^{'} y + c^{'} - r) = 0$

Let $C = (α, β)$ , then $C$ lies on $A C$ , therefore,

$a α + b β + c - p + k (a^{'} α + b^{'} β + c^{'} - r) = 0$ but $u = q$ and $v = s$

So the line becomes $q - p + k (s - r) = 0 ∴ k = \frac{p - q}{s - r}$

$\Rightarrow u - p + \frac{p - q}{s - r} (v - r) = 0 \Rightarrow u (r - s) - v (p - q) + p s - q r = 0$

$| \begin{matrix} u & v & 1 \\ p & r & 1 \\ q & s & 1 \end{matrix} | = 0$ .

We can write the given equation as $a (x + y - 1) + b (2 x + 3 y - 1) = 0$ . Clearly, both $a$ and $b$ cannot be zero at the same time. Let $a \neq 0$ , then

$x + y - 1 + \frac{b}{a} (2 x + 3 y - 1) = 0$ . This line passes through the point of intersection of $x + y - 1 = 0$ and $2 x + 3 y - 1 = 0$ , i.e. $x = 2, y = - 1$ .

Thus, all given straight lines pass through the fixed point $(2, - 1)$ .

Given that $a l + b m + c n = 0 => n = - \frac{a l + b m}{c}$ . Putting this in the equation of line

$l x + m y - \frac{a l + b m}{c} = 0 \Rightarrow l (c x - a) + m (c y - b) = 0$

Clearly, both $l$ and $m$ both can be zero because in the equation of straight line coeff. of both $x$ and $y$ cannot be zero.

Thus, the give equation represents straight lines which passes through the intersection of lines $c x - a = 0$ and $c y - b = 0$ i.e. through the point $(\frac{a}{c}, \frac{b}{c})$ .

Let $O$ the point of intersections of the lines be origin then we can represent the equations as $y = m_{r} x, r = 1, 2, 3, \dots, n$ .

Let the variable line be $y = m x + c$ . Solving gives us $x = \frac{c}{m_{e} - m}$ and $y = \frac{m_{r} c}{m_{r} - m}$

$O A_{r}^{2} = \frac{c^{2}}{(m_{r} - m)^{2}} (1 + m_{r}^{2}) \Rightarrow O A_{r} = | \frac{c}{M_{r} - m} | \sqrt{1 + m_{r}^{2}} \Rightarrow \frac{1}{O A_{r}} = | m_{r} - m | . \frac{1}{1 + m_{r}^{2}}$

$\Rightarrow \sum_{(} i = 1)^{n} \frac{1}{O A_{i}} = \pm \frac{m_{r} - m}{c} . \frac{1}{1 + m_{r}^{2}} = k$

Thus, $y = m x + c$ passes through the fixed point
$(\frac{1}{k} \sum \pm \frac{1}{\sqrt{1 + m_{r}^{2}}}, \frac{1}{k} \sum \pm \frac{m_{r}}{\sqrt{1 + m_{r}^{2}}})$ .

$Δ = | \begin{matrix} 4 & 7 & - 9 \\ 5 & - 8 & 15 \\ 9 & - 1 & 6 \end{matrix} | = 0$ .

Let $A B C$ be a triangle with vertices $A (x_{1}, y_{1}), B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ .

The median through $A$ will pass through $A (x_{1}, y_{1})$ and $D (\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2})$ . The median through $B$ will pass through $B$ and $E (\frac{x_{1} + x_{3}}{2}, \frac{y_{1} + y_{3}}{2})$ . The median through $C$ will pass through $C$ and $F (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

Equation of $A D$ is given by $y - y_{1} = \frac{\frac{y_{2} + y_{3}}{2} - y_{1}}{\frac{x_{2} + x_{3}}{2} - x_{1}} (x - x_{1})$

$\Rightarrow (2 y_{1} - y_{2} - y_{3}) x - (2 x_{1} - x_{2} - x_{3}) y - x_{1} (2 y_{1} - y_{2} - y_{3}) + y_{1} (2 x_{1} - x_{2} - x_{3}) = 0$

Similarly we can find the equation of $B E$ and $C F$ .We find that sum of equations is $0 = 0$ . Thus, medians are concurrent.

By trial we observe that $(p + q) x + (p + q) y - (p - q) + (p - q) x - (p - q) y - (p + q) + 2 [p x + q y - p] = 0$

Thus, first three lines are concurrent. We also see that

$(p - q) x - (p - q) y - (p + q) + p x + q y - p + q x + p y + q = 0$ .

Thus, last three lines are concurrent making all four lines concurrent.

The lines are concurrent, therefore, $| \begin{matrix} p_{1} & q_{1} & - 1 \\ p_{2} & q_{2} & - 1 \\ p_{3} & q_{3} & - 1 \end{matrix} | = 0 \Rightarrow | \begin{matrix} p_{1} & q_{1} & 1 \\ p_{2} & q_{2} & 1 \\ p_{3} & q_{3} & 1 \end{matrix} | = 0$ , and hence, the three points are concurrent.

Subtract the first equation from the second equation $(x + 2 y) - (x - 4 y) = 9 - 3 \Rightarrow 6 y = 6 \Rightarrow y = 1$

Substitute into $x + 2 y = 9, x + 2.1 = 9 \Rightarrow x = 7$

So the intersection point is $(7, 1)$ .

Now substituting into the line $m x + 2 y + 5 = 0 \Rightarrow m .7 . + 2.1 + 5 = 0 \Rightarrow 7 m + 7 = 0 \Rightarrow m = - 1$ .

Subtracting yields $y (t_{1} - t_{2}) = a (t_{1}^{2} - t_{2}^{2}) \Rightarrow y = a (t_{1} + t_{2}) \Rightarrow x = a t_{1} t_{2}$ .

From $x + y = 3$ , we get $x = 3 - y$ . Substitute into the second equation, $2 (3 - y) - 3 y = 1 \Rightarrow y = 1$

Then $x = 3 - 1 = 2$ . So the point of intersection is $(2, 1)$ .

Now the line is $\frac{x}{a} + \frac{y}{b} = 1$ and it passes through $(2, 1)$ :

$\Rightarrow \frac{2}{a} + \frac{1}{b} = 1$

Rewrite the line: $\frac{x}{a} + \frac{y}{a} = 1 \Rightarrow y = b - \frac{b}{a} x$ . So slope $= - \frac{b}{a}$ .

Given it is parallel to $y = x - 6$ , whose slope is $1$ :

$- \frac{b}{a} = 1 \Rightarrow b = - a$

Substitute into the earlier equation: $\frac{2}{a} + \frac{1}{- a} = 1 \Rightarrow a = 1$ . Then $b = - 1$ .

Intersection of $x = y$ and $y = 2 x : x = 2 x \Rightarrow x = 0, y = 0$

Intersection of $x = y$ and $y = 3 x + 4 : x = 3 x + 4 \Rightarrow - 2 x = 4 \Rightarrow x = - 2, y = - 2$

Intersection of $y = 2 x$ and $y = 3 x + 4 : 2 x = 3 x + 4 \Rightarrow - x = 4 \Rightarrow x = - 4, y = - 8$

Vertices: $(0, 0), (- 2, - 2), (- 4, - 8)$

Area $= \frac{1}{2} | 0 (- 2 + 8) + (- 2) (- 8 - 0) + (- 4) (0 + 2) | = 4$ .

From $3 x - 4 y + 4 a = 0$ and $2 x - 3 y + 4 a = 0 : 3 x - 4 y = - 4 a, 2 x - 3 y = - 4 a \Rightarrow 6 x - 8 y = - 8 a, 6 x - 9 y = - 12 a$

$=> y = 4 a$ . Substituteing, $2 x - 3 (4 a) = - 4 a \Rightarrow 2 x - 12 a = - 4 a \Rightarrow 2 x = 8 a \Rightarrow x = 4 a$ . Point: $(4 a, 4 a)$ .

From $2 x - 3 y + 4 a = 0$ and $5 x - y + a = 0 : 2 x - 3 y = - 4 a, 5 x - y = - a$

From second: $y = 5 x + a$ . Substituting: $2 x - 3 (5 x + a) = - 4 a \Rightarrow 2 x - 15 x - 3 a = - 4 a \Rightarrow - 13 x = - a => x = \frac{a}{13}$

$y = 5 \frac{a}{13} + a = \frac{18 a}{13}$ . Point: $(\frac{a}{13}, \frac{18 a}{13})$ .

From $3 x - 4 y + 4 a = 0$ and $5 x - y + a = 0$ . Point: $(0, a)$

Area $= \frac{1}{2} | 4 a (\frac{18 a}{13} - a) + \frac{a}{13} (a - 4 a) + 0 (4 a - \frac{18 a}{13}) | = \frac{17 a^{2}}{26}$ .

Intersection points: $y = m_{1} x$ and $y = m_{2} x$ give $(0, 0)$

$y = c$ with $y = m_{1} x$ gives $(\frac{c}{m_{1}}, c)$

$y = c$ with $y = m_{2} x$ gives $(\frac{c}{m_{2}}, c)$

So triangle vertices are $O (0, 0), A (\frac{c}{m 1}, c)$ , $B (c / m_{2}, c)$ .

Take $A B$ as base: Base length: $c * | \frac{1}{m_{1}} - \frac{1}{m - 2} | a b s ((1 / m_{1}) - (1 / m_{2})) = c . | \frac{m_{2} - m 2}{m_{1} m_{2}} |$

Height $= c$ . Area $= \frac{1}{2} . c . c . | \frac{m_{2} - m 1}{m_{1} m_{2}} |$

Now $m_{1}, m_{2}$ are roots of: $x^{2} + (\sqrt{3} + 2) x + (\sqrt{3} - 1) = 0$

Sum: $m_{1} + m_{2} = - s q r t 3 - 2$ . Product: $m_{1} m_{2} = \sqrt{3} - 1$

Difference: $| m_{2} - m_{1} | = \sqrt{(m_{1} + m_{2})^{2} - 4 m_{1} m_{2}} = \sqrt{11}$

So area: $Δ = \frac{c^{2}}{2} . \frac{\sqrt{11}}{\sqrt{3} - 1}$ .

Line perpendicular to $4 x + 7 y + 13 = 0$ is given by $7 x - 4 y + k = 0$ . Since $P (- 8, 12)$ lies on this perpendicular, therefore,

$k = 104$ . So the equation of the perpendicular is $7 x - 4 y + 104 = 0$ . This line intersects with the given line.

Solving both the equation we find the coordinate of the foot of the perpendicular as $(- 12, 5)$ .

Slope: $m = \frac{4 - 2}{5 - (- 1)} = \frac{1}{3}$ . Equation using point $P : y - 2 = \frac{1}{3} (x + 1)$

$x - 3 y + 7 = 0$

Let foot of perpendicular be $F (x, y)$ .

Since $A F$ is perpendicular to $P Q$ , slope of $P Q$ is $\frac{1}{3}$ , so slope of $A F$ is $- 3$ .

Equation of $A F$ through $A (1, 0) : y - 0 = - 3 (x - 1) \Rightarrow y = - 3 x + 3$

Now solve with line $P Q : x - 3 y + 7 = 0$

Substitute $y : x - 3 (- 3 x + 3) + 7 = 0$

$x = \frac{1}{5}, y = - 3 \frac{1}{5} + 3 = - \frac{3}{5} + \frac{15}{5} = \frac{12}{5}$ .

Line perpendicular to $x + 3 y = 3$ is $3 x - y = k$ but it passes through origin so $k = 0$ . Now intersection of $x + 3 y = 3$ and $3 x - y = 0$ is $(\frac{3}{10}, \frac{9}{10})$ .

Line perpendicular to $2 x + 3 y = 5$ is $3 x - 2 y = k^{'}$ but it passes through origin so $k^{'} = 0$ . Now intersection of the two line is $(\frac{10}{13}, \frac{15}{13})$ .

Now line passing through these points is given by $33 x - 61 y + 45 = 0$ .

Let $(h, r)$ be the foot of the perpendicular from $(x_{1}, y_{1})$ to the line.

Since $(h, r)$ lies on the line: $l h + m r + n = 0 \dots (1)$

Slope of given line is $- \frac{l}{m}$ . So slope of perpendicular is $\frac{m}{l}$ .

Hence, slope of line joining $(x_{1}, y_{1})$ and $(h, r)$ is $\frac{m}{l}$ :

$\frac{y_{1} - r}{x_{1} - h} = \frac{m}{l}$ . So: $l (y_{1} - r) = m (x_{1} - h)$

$l y_{1} - l r = m x_{1} - m h \dots (2)$

Rearranging (2): $m h - l r = m x_{1} - l y_{1} \dots (3)$

Now multiplying (1) by $m : l m h + m^{2} r + m n = 0 \dots (4)$

Multiplying (1) by $l : l^{2} h + l m r + l n = 0 \dots (5)$

Now solving for $h$ and $r$ using (3),(4),(5).

$h = x_{1} - \frac{l (l x_{1} + m y_{1} + n)}{l^{2} + m^{2}}, r = y_{1} - \frac{m (l x_{1} + m y_{1} + n)}{l^{2} + m^{2}}$

Hence: $x_{1} - h = \frac{l (l x_{1} + m y_{1} + n)}{l^{2} + m^{2}}, y_{1} - r = \frac{m (l x_{1} + m y_{1} + n)}{l^{2} + m^{2}}$

So dividing: $\frac{x_{1} - h}{l} = \frac{y_{1} - r}{m} = \frac{l x_{1} + m y_{1} + n}{l^{2} + m^{2}}$ .

Let image of $P$ be $P^{'} (x, y)$ . Let the foot of perpendicular be $F$ from $P$ to the line.

Slope of given line is $- \frac{4}{7}$ , so perpendicular slope is $\frac{7}{4}$ .

Equation of perpendicular through $(- 8, 12) : y - 12 = \frac{7}{4} (x + 8)$

$4 y - 48 = 7 x + 56 \Rightarrow 7 x - 4 y + 104 = 0$

Now solving with $4 x + 7 y + 13 = 0$ give us the foot foot as $F (- 12, 5)$ .

Since $F$ is midpoint of $P (- 8, 12)$ and $P^{'} (x, y)$ :

$x_{F} = \frac{x + (- 8)}{2} = - 12 \Rightarrow x = - 16$

$y_{F} = \frac{y + 12}{2} = 5 \Rightarrow y = - 2$ .

Let $A (2, 1)$ and its image be $A^{'} (5, 2)$ . The mirror line is the perpendicular bisector of segment $A A^{'}$ .

Midpoint of $A A^{'}$ is $M = (\frac{2 + 5}{2}, \frac{1 + 2}{2}) = (\frac{7}{2}, \frac{3}{2})$

Slope of $A A^{'}$ is $m = \frac{2 - 1}{5 - 2} = \frac{1}{3}$

So slope of mirror line is negative reciprocal is $- 3$ .

Equation of line through $M (\frac{7}{2}, \frac{3}{2})$ with slope $- 3$ is given by $y - \frac{3}{2} = - 3 (x - \frac{7}{2}) \Rightarrow 3 x + y = 12$ .

The point of intersection of the lines $3 x + 2 y = 0$ and $x - 2 y = 0$ is $O (0, 0)$ . Equation of the line passing through $O (0, 0)$ and $(1, 1)$ is given by $x = y$ .

Given that $5 x - y = 9 \Rightarrow y = 5 x - 9$ . Putting it in the line $x + 6 y = 8 \Rightarrow x + 30 x - 54 = 8 \Rightarrow x = 2 \Rightarrow y = 1$ .

Equation of the line passing through $(2, 1)$ and $(2, - 2)$ is given by $x = 2$ .

From first equation, $y = 1 - 2 x \Rightarrow x + 3 (1 - 2 x) - 2 = 0 \Rightarrow x = \frac{1}{5}$

Then $y = 1 - 2 \frac{1}{5} = \frac{3}{5}$ . So intersection point is $P (\frac{1}{5}, \frac{3}{5})$ .

Let required line cut axes at $(a, 0)$ and $(0, b)$ .

Equation in intercept form is given by $\frac{x}{a} + \frac{y}{b} = 1$

Since it passes through $P (\frac{1}{5}, \frac{3}{5}) \Rightarrow \frac{1}{5 a} + \frac{3}{5 b} = 1$

$\Rightarrow \frac{1}{a} + \frac{3}{b} = 5$

Area of triangle formed with axes is given by $1 / 2. a . b = \frac{3}{8} \Rightarrow a b = \frac{3}{4}$

$\Rightarrow \frac{b + 3 a}{a b} = 5 \Rightarrow \frac{b + 3 a}{\frac{3}{4}} = 5 \Rightarrow b + 3 a = \frac{15}{4}$

$\Rightarrow b = \frac{15}{4} - 3 a \Rightarrow a (\frac{15}{4} - 3 a) = \frac{3}{4} \Rightarrow 4 a^{2} - 5 a + 1 = 0$

$\Rightarrow a = 1$ or $a = \frac{1}{4}$

If $a = 1$ , then $b = \frac{3}{4}$ . If $a = \frac{1}{4}$ , then $b = 3$ .

Hence, equations are $x + \frac{y}{\frac{3}{4}} = 1 \Rightarrow 3 x + 4 y - 3 = 0$ and $\frac{x}{\frac{1}{4}} + \frac{y}{3} = 1 \Rightarrow 12 x + y - 3 = 0$ .

Figure 9.20.

Let $A B$ is $2 x - y + 1 = 0 \Rightarrow y = 2 x + 1$ . Let $A D$ is $x + 3 y - 10 = 0$

Figure 9.21.

Substitute $y = 2 x + 1$ into $A D : x + 3 (2 x + 1) - 10 = 0 \Rightarrow 7 x - 7 = 0 \Rightarrow x = 1, y = 3$

So $A = (1, 3)$ and $C = (- 1, - 2)$ Midpoint of $A C : M = (\frac{1 + (- 1)}{2}, \frac{3 + (- 2)}{2}) = (0, \frac{1}{2})$ .

So $M$ is midpoint of $B D$ also Let direction of $A B = (1, 2)$ . Let direction of $A D = (3, - 1)$ .

So $B = (1, 3) + t (1, 2) = (1 + t, 3 + 2 t)$ and $D = (1, 3) + s (3, - 1) = (1 + 3 s, 3 - s)$

Midpoint condition: $1 + t \frac{1 + 3 s}{2} = 0 - > t + 3 s = - 2 \Rightarrow (\frac{3 + 2 t + 3 - s}{}) = \frac{1}{2} \Rightarrow 2 t - s = - 5$

$\Rightarrow t = - \frac{17}{7}, s = \frac{1}{7}$ . So $B = (- \frac{10}{7}, - \frac{13}{7})$ and $D = (\frac{10}{7}, \frac{20}{7})$ .

Slope of $A C$ is $\frac{- 2 - 3}{- 1 - 1} = \frac{5}{2}$ and slope of $B D$ is $(\frac{\frac{20}{7} + \frac{13}{7}}{\frac{10}{7} + \frac{10}{7}}) = \frac{33}{20}$

Equation of $A B$ is $y - 3 = \frac{5}{2} (x - 1) \Rightarrow 5 x - 2 y + 1 = 0$

Equation of $B D$ is $33 x - 20 y + 10 = 0$ .

From $2 x - y - 5 = 0 \Rightarrow y = 2 x - 5$

Substituting into $3 x - y - 6 = 0$ gives us $3 x - (2 x - 5) - 6 = 0 \Rightarrow x = 1$

Now $y = 2.1 - 5 = - 3$ . So intersection point is $(1, - 3)$ .

$4 x - y - 7 = 0$ . Substitute $(1, - 3)$ gives us $4 (1) - (- 3) - 7 = 4 + 3 - 7 = 0$

Since it satisfies the third equation, all three lines are concurrent at $(1, - 3)$ .

Since the point lies on the $y$ -axis, $x = 0$ .

Substituting $x = 0$ in first line gives us $(2 m + 3) y + m + 6 = 0 \Rightarrow (2 m + 3) y = - (m + 6) \Rightarrow y = - \frac{m + 6}{2 m + 3}$

Substituting $x = 0$ in second line gives us $(m - 1) y + m - 9 = 0 \Rightarrow (m - 1) y = 9 - m \Rightarrow y = \frac{9 - m}{m - 1}$

Equating both values of $y$ gives us $- \frac{m + 6}{2 m + 3} = \frac{9 - m}{m - 1} \Rightarrow m = 21, - 1$ .

Substituting $y = x + 1$ into $2 x + y = 16$ gives us $2 x + (x + 1) = 16$

Simplifying gives us $3 x + 1 = 16$ . Solving gives us $3 x = 15$ so $x = 5$ .

Finding $y$ gives us $y = 5 + 1 = 6$

So the point of intersection being obtained is $(5, 6)$

Substituting $(5, 6)$ into $y = m x - 4$ gives us $6 = 5 m - 4 \Rightarrow m = 2$ .

Multiplying the first equation by $b$ is giving us $a b x + a^{2} b y + b = 0$

Multiplying the second equation by $a$ is giving us $a b x + a b^{2} y + a = 0$

Subtracting gives us $a^{2} b y - a b^{2} y + b - a = 0 \Rightarrow a b (a - b) y + (b - a) = 0$

$\Rightarrow (a - b) (a b y - 1) = 0$ . So either $a = b$ or $a b y = 1$ .

If $a = b$ , then at least two constants are already being equal.

If $a b y = 1$ , then $y = \frac{1}{a b}$ .

Substituting $y = \frac{1}{a b}$ into $a x + a^{2} y + 1 = 0$ gives us $a x + a^{2} \frac{1}{a b} + 1 = 0$

Simplifying gives us $a x + \frac{a}{b} + 1 = 0$

So $x$ is $x = - (\frac{\frac{a}{b} + 1}{a})$ .

Substituting the same point into the third equation $c x + c^{2} y + 1 = 0$ gives us a condition relating $a, b, c$

After simplification it gives us $(a - c) (b - c) = 0$

So either $a = c$ or $b = c$ . Therefore, at least two of $a, b, c$ are equal.

Equating $m_{1} x + c_{1} = m_{2} x + c_{2}$ gives us $(m_{1} - m_{2}) x = c_{2} - c_{1}$

So $x$ is $x = \frac{c_{2} - c_{1}}{m_{1} - m_{2}}$ . Finding $y$ gives us $y = \frac{m_{1} (c_{2} - c_{1})}{m - 1 - m_{2}} + c_{1}$

The third line also passes through this so $\frac{m_{1} (c_{2} - c_{1})}{m_{1} - m_{2}} + c_{1} = \frac{m_{3} (c_{2} - c_{1})}{m_{1} - m_{2}} + c_{3}$

$\Rightarrow m_{1} (c_{2} - c_{3}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0$ .

Multiplying $(b + c) x + a y + 1 = 0$ by $b$ gives us $b (b + c) x + a b y + b = 0$

Multiplying $(c + a) x + b y + 1 = 0$ by $a$ gives us $a (c + a) x + a b y + a = 0$

Subtracting the equations gives us $b (b + c) x - a (c + a) x + b - a = 0$

$\Rightarrow (b - a) ((a + b + c) x - 1) = 0$

So either $a = b$ or $(a + b + c) x = 1$ .

If $a = b$ , then two of the lines are already being identical in structure and concurrency is satisfied.

Assuming $a \neq b$ , finding $x$ gives us $x = \frac{1}{a + b + c}$

Substituting $x = \frac{1}{a + b + c}$ into the first equation gives us $\frac{b + c}{a + b + c} + a y + 1 = 0$

Solving for $y$ yields $a y = - 1 - \frac{b + c}{a + b + c}$

So $y = - \frac{a + b + c + b + c}{a (a + b + c)} = - \frac{a + 2 b + 2 c}{a (a + b + c)}$

Substituting this point into the third equation $(a + b) x + c y + 1 = 0$ yields $\frac{a + b}{a + b + c} + c \frac{- (a + 2 b + 2 b)}{a (a + b + c)} + 1 = 0$

Multiplying by $(a + b + c)$ yields $(a + b) - \frac{c (a + 2 b + 2 c)}{a} + (a + b + c) = 0$

Simplifying yields an identity equal to zero. So the same point is satisfying all three equations.

We consider a triangle with vertices $A (x_{1}, y_{1}), B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ .

We take the perpendicular bisector of side $A B$ . The midpoint of $A B$ is $M_{1} (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ and the slope of $A B$ is $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

So the slope of its perpendicular bisector is $- \frac{x_{2} - x_{1}}{y_{2} - y_{1}}$ .

Hence, the equation of the perpendicular bisector of $A B$ is $y - \frac{y_{1} + y_{2}}{2} = - \frac{x_{2} - x_{1}}{y_{2} - y_{2}} (x - \frac{x_{1} + x_{2}}{2})$

Similarly we form the perpendicular bisector of side $B C$ . Its midpoint is $M_{2} (\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2})$ and its equation is $y - \frac{y_{2} + y_{3}}{2} = - \frac{x_{3} - x_{2}}{y_{3} - y_{2}} (x - \frac{x_{2} + x_{3}}{2})$

Now we solve these two equations simultaneously and obtain a point $(h, k)$ .

This point $(h, k)$ satisfies both equations, so it lies on the perpendicular bisectors of $A B$ and $B C$ . Hence it is equidistant from $A$ and $B$ , and also from $B$ and $C$ .

Thusm we get $P A = P B$ and $P B = P C$ so we conclude $P A = P C$ .

This shows that the point $(h, k)$ also lies on the perpendicular bisector of $A C$ .

Let $A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})$ be the vertices of a $△ A B C$ .

Figure 9.22.

Let $P (x, y)$ lie on the perpendicular bisector of $A B$ , then $P A = P B$

$\sqrt{(x - x_{1})^{2} + (y - y_{1})^{2}} = \sqrt{(x - x_{2})^{2} + (y - y_{2})^{2}}$

$\Rightarrow 2 x (x_{2} - x_{1}) + 2 y (y_{2} - y_{1}) = x_{2}^{2} + y_{2}^{2} - x_{1}^{2} - y_{1}^{2}$

This is the perpendicular bisector of $A B$

Similarly for $B C$ , $2 x (x_{3} - x_{2}) + 2 y (y_{3} - y_{2}) = x_{3}^{2} + y_{3}^{2} - x_{2}^{2} - y_{2}^{2}$

The two linear equations intersect at a unique point and that point is equidistant from $A, B$ , and $C$ ie. $P A = P B = P C$

Therefore, the perpendicular bisectors of a triangle are concurrent.

Given equation is $x (1 + λ) + y (2 - λ) + 5 = 0$ , which can be written as $x + 2 y - 5 + λ (x - y) = 0$

The above equation represents two lines $x + 2 y - 5 = 0$ and $x - y = 0$ to be concurrent. Solving the two equations we find the fixed point as $(- \frac{5}{3}, \frac{- 5}{3})$ .

$x (a + 2 b) + y (a - 3 b) = a - b$ can be rewritten as $a (x + y - 1) + b (2 x - 3 y + 1) = 0$ ,which represents two equations $x + y - 1 =$ and $2 x - 3 y + 1 = 0$ , which are concurrent.

Solving the two equation yields the fixed point $(\frac{2}{5}, \frac{3}{5})$ , which is independent of $a$ and $b$ .

Solving first two equation gives $A = (0, 0)$ , Solving first and last gives $B = (20, 15)$ and solving last two gives $C = (- 36, 15)$

Figure 9.23.

If $G$ be the centroid then $G = (\frac{0 + 26 - 26}{3}, \frac{0 + 15 + 15}{3}) = (- \frac{16}{3}, 10)$ .

$a = B C = 39, b = C A = 56, c = A B = 39$

Using the formula for the incenter we have $I = (- 1, 8)$ .

Let $A = (0, 0), B = (2, - 1)$ and $C = (- 1, 3)$ . Let $A L ⊥ B C$ then equation is $3 x = 4 y$ .

Figure 9.24.

Let $B M ⊥ A C$ , then equation of $B M$ is $x - 3 y = 5$ . Solving the equation of two perpendiculars we get orthocenter as $(- 4, - 3)$ .

Consider a triangle $A B C$ with sides $A B : 3 x - 2 y = 6, B C : 3 x + 4 y = - 12$ and $A C : 3 x - 8 y = - 12$ .

Solving first and last we get $A = (4, 3)$ , and solving first two gives us $B = (0, - 3)$ .

Let $A L ⊥ B C$ then equation of $A L$ is given by $4 x - 3 y - 7 = 0$ , and if $B M ⊥ A C$ then equation of $B M$ is given by $8 x + 3 y + 9 = 0$ .

Solving $A L$ and $B M$ gives us the orthocenter as $H = (- \frac{1}{6}, - \frac{23}{9})$ .

Let $A B C$ be the given triangle and $B = (3, - 1)$ and $C = (- 2, 3)$ . Let $H$ be the orthocenter of the $△ A B C$ .

Given that $H = (0, 0)$ . Since $A B$ passes through $B (3, - 1)$ and is perpendicular to the line $C H$ equation of $A B$ is $2 x - 3 y = 9$ . Similarly equation of $A C$ is $3 x - y = - 9$ .

Solving the two equations gives us $A = (- \frac{36}{7}, - \frac{45}{7})$ .

Consider a $△ A B C$ such that $A B$ is $y = m_{1} x$ and $A C$ is $y = m_{2} x$ . We also let equation of $B C$ as $l x + m y = 1$ .

Clearly, $A$ will be $(0, 0)$ .

Since $A H ⊥ B C ∴ \frac{b}{a} \frac{- l}{m} = - 1 \Rightarrow \frac{l}{m} = \frac{a}{b} = k$ (say)

Coordinates of $B$ and $C$ are $(\frac{1}{l + m m_{1}}, \frac{m_{1}}{l + m m_{1}})$ and $(\frac{1}{l + m m_{2}}, \frac{m_{2}}{l + m m_{2}})$ .

Equation of perpendicular bisectors through $B$ and $C$ are

$x + m_{2} y = \frac{1 + m_{1} m_{2}}{l + m m_{1}}$ and $x + m_{1} y = (1 + m_{1} m_{2}) / (l + m m_{2})$

Thus, coordinates of $H$ are $y = \frac{(1 + m_{1} m_{2}) m_{1}}{l^{2} + l m (m_{1} + m) 2 + m^{2} m_{1} m_{2}))}$

$m_{1} + m_{2} = - \frac{2 h}{b}$ and $m_{1} m_{2} = \frac{a}{b}$

Thus, $y = \frac{(a + b) m}{b l^{2} - 2 h l m - a m^{2}} \Rightarrow \frac{m}{b} = \frac{b l^{2} - 2 h l m + a m^{2}}{a + b}$

$\Rightarrow k = \frac{a + b}{a b (a + b - 2 h)}$

Thus, equation of $B C$ is $l x + m y = 1 \Rightarrow a b x + b k y = 1 \Rightarrow k (a x + b y) = 1$

$\Rightarrow (a x + b y) (a + b) = a b (a + b - 2 h)$ .

Consider the $△ A B C$ such that equation of $A B$ is $p x + q y + r = 0$ and that of $A C$ is $l x + m y + n = 0$ .

Equation of any line passing through these lines is given by $p x + q y + r + k (l x + m y + n) = 0$ . Slope of this line is $- \frac{p + k l}{q + k m}$ .

Slope of $B C$ is $- \frac{a}{b}$ . Let $A D$ be the perpendicular through $A$ on $B C$ , then $A D ⊥ B C$ .

$\Rightarrow - \frac{p + k l}{q + k m} . - \frac{a}{b} = - 1 \Rightarrow k = \frac{c p + b q}{a l + b m}$

Thus, equation of $A D$ is $\frac{p x + q y + r}{a p + b q} = \frac{l x + m y + n}{a l + n b}$ .

Let the equations of the sides $B C, C A$ and $A B$ of the $△ A B C$ are $L_{1} == x \cos θ_{1} + y \sin θ_{1} - p_{1} = 0, x \cos θ_{2} + y \sin θ_{2} - p_{2} = 0$ and $x \cos θ_{3} + y \sin θ_{3} - p_{3} = 0$ .

Let $A D$ and $B E$ are perpendiculars through $A$ and $B$ on opposite sides.

Equation of $A D$ is given by $L_{2} + k L_{3} = 0 \Rightarrow x (\cos θ_{2} + \cos θ_{3}) + y (\sin θ_{2} + \sin θ_{3}) - (p_{2} + k p_{3}) = 0$

Slope of $A D$ is $- \frac{\cos θ_{2} + \cos θ_{3}}{\sin θ_{2} + \sin θ_{3}} = m_{1}$

Slope of $B C$ is $- \frac{\cos θ_{1}}{\sin θ_{1}}$ . Product of these two slopes would be $- 1$ .

$\Rightarrow k = - \frac{\cos (θ_{1} - θ_{2})}{\cos (θ_{1} - θ_{3})}$

Thus, equation of $A D$ becomes $L_{2} - \frac{\cos (θ_{1} - θ_{2})}{\cos (θ_{1} - θ_{3})} L_{3} = 0 \Rightarrow L_{1} \cos (θ_{1} - θ_{3}) = L_{3} \cos (θ_{1} - θ_{2})$

And we proceed similarly for another perpendiculars to obtain the desired equation.

$A = (0, 0)$ from $3 x - 4 y = 0$ and $5 x + 12 y = 0$

$B = (20, 15)$ from $3 x - 4 y = 0$ and $y = 15$

$C = (- 36, 15)$ from $5 x + 12 y = 0$ and $y = 15$

Centroid is $G = (\frac{0 + 20 - 36}{3}, \frac{0 + 15 + 15}{3}) = (- \frac{16}{3}, 10)$

Length of the sides are, $a = | B C | = 56, b = | C A | = 39, c = | A B | = 25$

Incenter is $I = (\frac{a x_{A} + b x_{B} + c x_{C}}{a + b + c}, \frac{a y_{A} + b y_{B} + c y_{C}}{a + b + c}) = (\frac{0 + 780 - 900}{120}, \frac{0 + 585 + 375}{120}) = (- 1, 8)$ .

Let $A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})$ .

$A B^{2} = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} . A C^{2} = (x_{1} - x_{3})^{2} + (y_{1} - y_{3})^{2}$ , and $B C^{2} = (x_{2} - x_{3})^{2} + (y_{2} - y_{3})^{2}$

Consider $A B^{2} + A C^{2} - B C^{2} = [(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}] + [(x_{1} - x_{3})^{2} + (y_{1} - y_{3})^{2}] - [(x_{2} - x_{3})^{2} + (y_{2} - y_{3})^{2}]$

$= (x_{1}^{2} - 2 x_{1} x_{2} + x_{2}^{2}) + (y_{1}^{2} - 2 y_{1} y_{2} + y_{2}^{2}) + (x_{1}^{2} - 2 x_{1} x_{3} + x_{3}^{2}) + (y_{1}^{2} - 2 y_{1} y_{3} + y_{3}^{2}) - (x_{2}^{2} - 2 x_{2} x_{3} + x_{3}^{2}) - (y_{2}^{2} - 2 y_{2} y_{3} + y_{3}^{2})$

$= 2 [(x_{1}^{2} - x_{1} x_{2} - x_{1} x_{3} + x_{2} x_{3}) + (y_{1}^{2} - y_{1} y_{2} - y_{1} y_{3} + y_{2} y_{3})]$

$= 2 [(x_{1} - x_{2}) (x_{1} - x_{3}) + (y_{1} - y_{2}) (y_{1} - y_{3})]$

Thus, $A B^{2} + A C^{2} - B C^{2} = 2 [(x_{1} - x_{2}) (x_{1} - x_{3}) + (y_{1} - y_{2}) (y_{1} - y_{3})]$

We know that if $∠ A$ is acute if $A B^{2} + A C^{2} - B C^{2} > 0$ and obtuse if $< 0$ and for right angle should be equal to zero.

Given lines are $L_{1} : 4 x - 3 y = 5, L_{2} : x - 2 y = 10, L_{3} : 7 x + y = 40$ , and $L_{4} : x + 3 y + 10 = 0$

Let $A$ be the intersection of $L_{1}$ and $L_{2}, B$ be the intersection of $L_{2}$ and $L_{3}, C$ be the intersection of $L_{3}$ and $L_{4}$ , and $D$ be the intersection of $L_{4}$ and $L_{1}$ .

Intersection of $L_{1}$ and $L_{2}$ is $4 x - 3 y = 5$ and $x - 2 y = 10$

From $x - 2 y = 10$ , we get $x = 10 + 2 y \Rightarrow 4 (10 + 2 y) - 3 y = 5 \Rightarrow 5 y = - 35 \Rightarrow y = - 7 \Rightarrow x = 10 + 2 (- 7) = - 4 \Rightarrow A$ is $(- 4, - 7)$

Similarly we find that other points are $B (6, - 2), C (\frac{13}{2}, - \frac{11}{2})$ , and $D (- 1, - 3)$

We find the slopes as $m_{A B} = \frac{- 2 + 7}{6 + 4} = \frac{1}{2}, m_{B C} = (\frac{\frac{- 11}{2} + 2}{\frac{13}{2} - 6}) = - 7, m_{C D} = (\frac{- 3 + \frac{11}{2}}{- 1 - \frac{13}{2}}) = - \frac{1}{3}$ , and $m_{A C} = \frac{- 7 + 3}{- 4 + 1} = \frac{4}{3}$

Now we check angle relations using: $\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$

We find that $∠ A + ∠ C = 180^{\circ}$ . Hence, the quadrilateral is cyclic.

Let the four sides of the quadrilateral taken in order be $L_{1} : a_{1} x + b_{1} y + c_{1} = 0, L_{2} : a_{2} x + b_{2} y + c_{2} = 0, L_{3} : a_{3} x + b_{3} y + c_{3} = 0$ , and $L_{4} : a_{4} x + b_{4} y + c_{4} = 0$ .

Let the vertices be $A = L_{1} \cap L_{2}, B = L_{2} \cap L_{3}, C = L_{3} \cap L_{4}, D = L_{4} \cap L_{1}$

A quadrilateral is cyclic iff opposite angles are supplementary i.e. $∠ A + ∠ C = 180^{\circ}$

For a line $a x + b y + c = 0$ , slope is $m = - \frac{a}{b}$

Angle between two lines $L_{1}$ and $L_{2}$ is $\tan θ = | \frac{a_{2} b_{1} - a_{1} b_{2}}{a_{1} a_{2} + b_{1} b_{2}} |$

$\Rightarrow \tan A = | \frac{a_{2} b_{1} - a_{1} b_{2}}{a_{1} a_{2} + b_{1} b_{2}} |$

$\Rightarrow \tan C = | \frac{a_{4} b_{3} - a_{3} b_{4}}{a_{3} a_{4} + b_{3} b_{4}} |$

For cyclic quadrilateral, $∠ A + ∠ C = 180^{\circ}$

$\Rightarrow \frac{a_{2} b_{1} - a_{1} b_{2}}{a_{1} a_{2} + b_{1} b_{2}} = \frac{a_{4} b_{3} - a_{3} b_{4}}{a_{3} a_{4} + b_{3} b_{4}}$ .

Show that the lines $2 x + 3 y + 19 = 0$ and $9 x + 6 y - 17 = 0$ cut the coordinate axes in concyclic points.

Let the given lines be $L_{1} : 2 x + 3 y + 19 = 0$ , and $L_{2} : 9 x + 6 y - 17 = 0$ .

For $L_{1}, x$ -intercept: put $y = 0 \Rightarrow 2 x + 19 = 0 \Rightarrow x = - \frac{19}{2}$ . So $A$ is $(- \frac{19}{2}, 0)$

$y$ -intercept: put $x = 0 \Rightarrow 3 y + 19 = 0 \Rightarrow y = - \frac{19}{3}$ . So $B$ is $(0, - \frac{19}{3})$

Similarly we find $C$ and $D$ for $L_{2}$ as $C (\frac{17}{9}, 0)$ and $D (0, \frac{17}{6})$

Now we check concyclicity using the condition that four points are concyclic if the angle subtended by the same chord is equal.

Consider chord $A C$ on the $x$ -axis.

Slope of $A B$ is $m_{A B} = \frac{- \frac{19}{3} - 0}{0 + \frac{19}{2}} = - \frac{2}{3}$

Slope of $B C$ is $m_{B C} = \frac{- \frac{19}{3} - 0}{0 - \frac{17}{9}} = \frac{57}{17}$

So angle at $B$ is angle between lines with slopes $- - \frac{2}{3}$ and $\frac{57}{17}$ .

Now slope of $A D$ is $m_{A D} = \frac{\frac{17}{6} - 0}{0 + \frac{19}{2}} = \frac{17}{57}$

Slope of $C D$ is $m_{C D} = \frac{\frac{17}{6} - 0}{0 - \frac{17}{9}} = - \frac{3}{2}$

So angle at $D$ is angle between slopes $\frac{17}{57}$ and $- \frac{3}{2}$ .

Now compute angle between two lines using: $\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$

For angle at $B$ is $\tan ∠ B = | \frac{\frac{57}{17} + \frac{2}{3}}{1 - \frac{2}{3} . \frac{57}{17}} | = 1$

For angle at $D$ is $\tan ∠ D = | \frac{- \frac{3}{2} - \frac{17}{57}}{1 - \frac{17}{57} . \frac{- 3}{2}} | = 1$

Thus $∠ B = ∠ D$ . Since equal angles subtend the same chord, the four points are concyclic.

Let $B (- 4, - 5)$ be a vertex of triangle $A B C$ .

$5 x + 3 y - 4 = 0$ is altitude from $A$ , so slope of $B C$ is $\frac{3}{5}$ . Equation of $B C$ through $B (- 4, - 5)$ is $y + 5 = \frac{3}{5} (x + 4) \Rightarrow 3 x - 5 y - 13 = 0$

$3 x + 8 y + 13 = 0$ is altitude from $C$ , so slope of $A B$ is $\frac{8}{3}$ .

Equation of $A B$ through $B (- 4, - 5)$ is $y + 5 = \frac{8}{3} (x + 4) \Rightarrow 8 x - 3 y + 17 = 0$

We find $A$ as intersection of $A B$ and altitude $5 x + 3 y - 4 = 0$

$13 x + 13 = 0 \Rightarrow x = - 1 \Rightarrow 5 (- 1) + 3 y - 4 = 0 \Rightarrow y = 3$ . So $A$ is $(- 1, 3)$ .

Similarly $C$ is $(1, - 2)$

Equation of $A C$ is $5 x + 2 y - 1 = 0$

So the sides are: $A B : 8 x - 3 y + 17 = 0, B C : 3 x - 5 y - 13 = 0$ , and $A C : 5 x + 2 y - 1 = 0$ .

Equation of the line perpendiculars to $5 x - y = 1$ is given by $x + 5 y = k$ . This will make an intercept of $k$ with $x$ -axis, and an intercept of $\frac{k}{5}$ with $y$ -axis.

Thus, area of triangle, which is given as $5$ , is $\frac{1}{2} . k . \frac{k}{5} = 25 \Rightarrow k = \pm 5 \sqrt{2}$ .

So the equation of the line is $x + 5 y = \pm 5 \sqrt{2}$ .

Clearly, $A = (6, 0)$ and $B = (0, 4)$ as given line is $\frac{x}{6} + \frac{y}{3} = 1$ .

We rewrite the line as $y = - \frac{2}{3} x + 4$ , so the slope of $A B$ is $- \frac{2}{3}$ .

A perpendicular line has slope $\frac{3}{2}$ , so we form the line through $(5, 5)$ as $y - 5 = \frac{3}{2} (x - 5) \Rightarrow y = \frac{3}{2} x - \frac{5}{2}$ .

Now we find point $C$ by setting $y = 0 : 0 = \frac{3}{2} x - \frac{5}{2}$ , so $x = \frac{5}{3}$ . Thus $C = (\frac{5}{3}, 0)$ .

We find point $D$ by setting $x = 0 : y = - \frac{5}{2}$ , so $D = (0, - \frac{5}{2})$ .

Now we find point $E$ by solving intersection of $2 x + 3 y = 12$ and $y = \frac{3}{2} x - \frac{5}{2}$ .

Thus, $x = 3$ , and $y = 2$ . So $E = (3, 2)$ .

We compute the area by splitting the quadrilateral $O C E B$ into two triangles $O C E$ and $O E B$ .

We first compute the area of triangle $O C E$ . Since $O = (0, 0)$ , we use the determinant shortcut: $Δ O C E = \frac{1}{2} | x_{C} y_{E} - x_{E} y_{C} | = \frac{5}{3}$ .

$Δ O E B = \frac{1}{2} | x_{E} y_{B} - x_{B} y_{E} | \Rightarrow \frac{1}{2} | 3.4 - 0.2 | = 6$ .

$◻ O C E B = \frac{5}{3} + 6 = \frac{23}{3}$ .

Figure 9.25.

Since the square is centered at the origin, we generate the remaining vertices by rotating the point $(1, 2)$ by $90^{\circ}$ repeatedly about the origin.

We rotate $(x, y)$ counterclockwise by $90^{\circ}$ using the rule $(x, y) \to (- y, x)$ .

So we obtain $A = (1, 2), B = (- 2, 1), C = (- 1, - 2)$ , and $D = (2, - 1)$ .

We first take line $A B$ . We compute its slope as $\frac{1 - 2}{- 2 - 1} = \frac{1}{3}$ .

So we write $y - 2 = \frac{1}{3} (x - 1)$ and simplify it to $x - 3 y + 5 = 0$ .

Similarly we find remaining sides to be $3 x + y + 5 = 0, x - 3 y - 5 = 0$ , and $3 x + y - 5 = 0$ .

We first find the slope of $A D$ as $\frac{6 - 2}{- 2 - 1} = - \frac{4}{3}$ .

Since $B C$ is perpendicular to $A D$ , we take the slope of $B C$ as $\frac{3}{4}$ .

So the line $B C$ passes through $D (- 2, 6)$ and we write $y - 6 = \frac{3}{4} (x + 2)$ x, which simplifies to $4 y - 24 = 3 x + 6$ , hence $3 x - 4 y + 30 = 0$ .

We note that the altitude in an equilateral triangle satisfies $h = \frac{\sqrt{3}}{2} s$ .

Figure 9.26. Figure for problem 231

We compute the altitude as $A D = \sqrt{(1 + 2)^{2} + (2 - 6)^{2}} = 5$ .

So the side length is $s = \frac{2 h}{\sqrt{3}} = \frac{10}{\sqrt{3}}$ .

Since $D$ lies on $B C$ , we take direction vector of $B C$ as $(4, 3)$ with magnitude $5$ , so the unit direction is $(\frac{4}{5}, \frac{3}{5})$ .

Half the side length is $\frac{s}{2} = \frac{5}{\sqrt{3}}$ .

So we move from $D$ to $B$ and $C$ using this direction $(\frac{5}{\sqrt{3}}) . (\frac{4}{5}, \frac{3}{5}) = (\frac{4}{\sqrt{3}}, \frac{3}{\sqrt{3}})$ .

Therefore, $B = (- 2 - \frac{4}{\sqrt{3}}, 6 - \frac{3}{\sqrt{3}})$ and $C = (- 2 + \frac{4}{\sqrt{3}}, 6 + \frac{3}{\sqrt{3}})$ .

Now we find line $A B$ using points $A (1, 2)$ and $B$ i.e. $y - 2 = m_{A B} (x - 1)$ where $m_{A B} = \frac{6 - \frac{3}{\sqrt{3}} - 2}{2 - \frac{4}{\sqrt{3}} - 1}$ .

Similarly we can find other sides.

We have one side $x - y = 0$ so its slope is $1$ . In an equilateral triangle, the angle between sides is $60^{\circ}$ . Hence slopes of the other sides satisfy

$m = \tan (45^{\circ} \pm 60^{\circ})$

So $m = \tan 105^{\circ} = - 2 - \sqrt{3}$ or $m = \tan - 15^{\circ} = 2 - \sqrt{3}$ .

Using the given vertex $(2 + \sqrt{3}, 5)$ and slope $2 - \sqrt{(} 3)$

$y - 5 = (2 - \sqrt{3}) (x - (2 + \sqrt{3}))$

Simplifying $y = (2 - \sqrt{3}) x + 6$

So second side is $y + (2 - \sqrt{3}) x = 6$ .

For the third side take slope $- 2 - \sqrt{3} \Rightarrow y - 5 = (- 2 - \sqrt{3}) (x - (2 + \sqrt{3}))$

$y + (2 + \sqrt{3}) x = 12 + 4 \sqrt{3}$ .

We have diagonal $8 x - 15 y = 0$ so its slope is $\frac{8}{15}$ . Hence the other diagonal has slope $- \frac{15}{8}$ .

In a square, diagonals bisect at right angles. Let the center be $(h, k)$ on $8 h - 15 k = 0$ .

Since $(1, 2)$ is a vertex, the midpoint lies on the line through $(1, 2)$ with slope $- \frac{15}{8}$

$k - 2 = - \frac{15}{8} (h - 1)$

Solving with $8 h - 15 k = 0$ gives $(h, k) = (\frac{16}{17}, \frac{120}{17})$ .

Slope of side is perpendicular to diagonal slope $\frac{8}{15}$ , so slope of side is $- \frac{15}{8}$ or $\frac{8}{15}$ rotated by $45^{\circ}$ . Thus, side slopes are $\frac{1}{8}$ and $- 8$ .

Through $(1, 2) \Rightarrow y - 2 = \frac{1}{8} (x - 1)$ gives $x - 8 y + 15 = 0$ and $y - 2 = - 8 (x - 1)$ gives $8 x + y - 10 = 0$ .

$5 y = 12 x + 6$ gives slope $\frac{12}{5}$ . $3 x = 4 y + 7$ gives slope $\frac{3}{4}$

Let required line have slope $m$ . For equal angles

$\frac{m - \frac{12}{5}}{1 + \frac{12}{5} m} = - \frac{m - \frac{3}{4}}{1 + \frac{3}{4} m}$

Solving gives $m = 1$ or $m = - 1$ .

Since these lines pass through $(4, 5)$ , therefore, $y - 5 = 1 (x - 4)$ gives $y = x + 1$ and $y - 5 = - 1 (x - 4)$ gives $y = - x + 9$ .

The two given lines are $7 x - y + 3 = 0$ and $x + y - 3 = 0$ . Solving them together gives $y = 7 x + 3$ and $y = 3 - x$ , so $7 x + 3 = 3 - x$ , which gives $x = 0$ and $y = 3$ . Thus the vertex of the triangle is $(0, 3)$ .

The angle bisectors are found from $\frac{7 x - y + 3}{5} = \pm (x + y - 3)$ . Solving gives the two bisectors $x - 3 y + 9 = 0$ and $3 x + y - 3 = 0$ .

The internal bisector is $3 x + y - 3 = 0$ , whose slope is $- 3$ . The base is perpendicular to this bisector, so its slope is $\frac{1}{3}$ .

Since the base passes through $(1, - 10)$ , its equation is $y + 10 = \frac{1}{3} (x - 1)$ . Simplifying gives $3 y + 30 = x - 1$ , hence the required equation is $x - 3 y - 31 = 0$ .

The three lines are $x \cos α + y \sin α = p_{1}, x \cos β + y \sin β = p_{2}$ , and $x \cos γ + y \sin γ = p_{3}$ .

The intersection point of the first two lines is obtained by solving $x \cos α + y \sin α = p_{1}$ and $x \cos β + y \sin β = p_{2}$ . Using determinants, this gives $x = \frac{p_{1} \sin β - p_{2} \sin α}{\sin (β - α)}$ and $y = \frac{p_{2} \cos α - p_{1} \cos β}{s i n (β - α)}$ .

Similarly, the other two vertices are obtained by cyclic permutation of $α, β, γ$ and $p_{1}, p_{2}, p_{3}$ .

The area of the triangle formed by three lines is given by $Δ = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$ .

Substituting the coordinates of the three intersection points and simplifying using the identity $\sin (A - B) = \sin A \cos B - \cos A \sin B$ , the expression is reduced.

The numerator becomes $(p_{1} \sin (γ - β) + p_{2} \sin (α - γ) + p_{3} \sin (β - α))^{2}$ and the denominator becomes $\sin (γ - β) \sin (α - γ) \sin (β - α)$ .

Thus, the area is $\frac{1}{2} \frac{(p_{1} \sin (γ - β) + p_{2} \sin (α - γ) + p_{3} \sin (β - α))^{2}}{| \sin (γ - β) \sin (α - γ) \sin (β - α) |}$ .

The line $L$ passes through $(1, 1)$ and $(2, 0)$ , so its slope is $m = \frac{0 - 1}{2 - 1} = - 1$ . Hence its equation is $y - 1 = - 1 (x - 1)$ , which simplifies to $y = - x + 2$ .

A line perpendicular to $L$ has slope $1$ . Passing through $(\frac{1}{2}, 0)$ , its equation is $y - 0 = 1 (x - \frac{1}{2})$ , so $y = x - \frac{1}{2}$ .

The three lines are $x = 0, y = - x + 2$ , and $y = x - \frac{1}{2}$ .

The intersection of $x = 0$ and $y = - x + 2$ is $(0, 2)$ . The intersection of $x = 0$ and $y = x - \frac{1}{2}$ is $(0, - \frac{1}{2})$ .

The intersection of $y = - x + 2$ and $y = x - \frac{1}{2}$ is found by solving $- x + 2 = x - \frac{1}{2}$ , which gives $2 x = \frac{5}{2}$ , so $x = \frac{5}{4}$ and $y = \frac{3}{4}$ .

Thus, the vertices of the triangle are $(0, 2), (0, - \frac{1}{2})$ , and $(\frac{5}{4}, \frac{3}{4})$ .

The base along the $y$ -axis has length $| 2 - (- \frac{1}{2}) | = \frac{5}{2}$ , and the perpendicular distance of $(\frac{5}{4}, \frac{3}{4})$ from the $y$ -axis is $\frac{5}{4}$ .

Hence, the area is $\frac{1}{2} . \frac{5}{2} . \frac{5}{4} = \frac{25}{16}$ .

The vertices of the triangle are $A (0, 0), B (8, 0)$ , and $C (4, 8)$ , and the given point is $P (9, 3)$ .

The side $A B$ has equation $y = 0$ . The foot of the perpendicular from $P (9, 3)$ to this line is clearly $(9, 0)$ .

The side $B C$ passes through $(8, 0)$ and $(4, 8)$ , so its slope is $m = \frac{8 - 0}{4 - 8} = - 2$ . Hence, its equation is $y = - 2 x + 16$ , or $2 x + y - 16 = 0$ .

The foot of the perpendicular from $(9, 3)$ to $2 x + y - 16 = 0$ is given by $x = 9 - 2 \frac{2.9 + 3 - 16}{2^{2} + 1^{2}}$ and $y = 3 - \frac{2.9 + 3 - 16}{2^{2} + 1^{2}}$ .

Since $2.9 + 3 - 16 = 5$ , this gives $x = 9 - \frac{10}{5} = 7$ and $y = 3 - \frac{5}{5} = 2$ , so the foot is $(7, 2)$ .

The side $C A$ passes through $(4, 8)$ and $(0, 0)$ , so its slope is $2$ . Its equation is $y = 2 x$ , or $2 x - y = 0$ .

The foot of the perpendicular from $(9, 3)$ to $2 x - y = 0$ is $x = 9 - 2 \frac{2.9 - 3}{2^{2} + (- 1)^{2}}$ and $y = 3 - \frac{(- 1) (29 - 3)}{2^{2} + (- 1)^{2}}$ .

Since $29 - 3 = 15$ , this gives $x = 9 - \frac{30}{5} = 3$ and $y = 3 + \frac{15}{5} = 6$ , so the foot is $(3, 6)$ .

Thus, the three feet are $(9, 0), (7, 2)$ , and $(3, 6)$ .

To check collinearity, the slope between $(9, 0)$ and $(7, 2)$ is $\frac{2 - 0}{7 - 9} = - 1$ , and the slope between $(7, 2)$ and $(3, 6)$ is $\frac{6 - 2}{3 - 7} = - 1$ , so the points lie on a straight line.

The equation of the line through $(9, 0)$ with slope $- 1$ is $y - 0 = - 1 (x - 9)$ , which simplifies to $x + y - 9 = 0$ .

The line is $\frac{x}{a} + \frac{y}{b} = 1$ , which can be written as $b x + a y - a b = 0$ .

The foot of the perpendicular from the origin $(0, 0)$ to the line $b x + a y - a b = 0$ is given by $α = - \frac{b (- a b)}{a^{2} + b^{2}}$ and $β = - \frac{a (- a \b)}{a^{2} + b^{2}}$ .

Thus, $α = \frac{a b^{2}}{a^{2} + b^{2}}$ and $β = \frac{a^{2} b}{a^{2} + b^{2}}$ .

Now $α^{2} + β^{2} = \frac{a^{2} b^{4} + a^{4} b^{2}}{(a^{2} + b^{2})^{2}} = \frac{a^{2} b^{2}}{a^{2} + b^{2}}$ .

Also $α + β = \frac{a b^{2} + a^{2} b}{a^{2} + b^{2}} = \frac{a b (a + b)}{a^{2} + b^{2}}$ .

Thus $(α^{2} + β^{2}) (α + β) = \frac{a^{3} b^{3} (a + b)}{a^{2} + b^{2})^{2}}$ .

Further $α β = \frac{a^{3} b^{3}}{(a^{2} + b^{2})^{2}}$ .

Hence, $(a + b) α β = \frac{a^{3} b^{3} (a + b)}{(a^{2} + b^{2})^{2}}$ .

Thus, $(α^{2} + β^{2}) (α + β) = (a + b) α β$ .

The given lines are $x = 0, y = 0, x + y = 1$ , and $6 x + y = 3$ .

The vertices of the quadrilateral are obtained by pairwise intersections. The origin $(0, 0)$ is one vertex. The intersection of $y = 0$ and $x + y = 1$ is $(1, 0)$ . The intersection of $x + y = 1$ and $6 x + y = 3$ is found by subtracting the equations, giving $5 x = 2$ , so $x = \frac{2}{5}$ and $y = \frac{3}{5}$ . The intersection of $6 x + y = 3$ and $x = 0$ is $(0, 3)$ .

Thus, the quadrilateral has vertices $(0, 0), (1, 0), (\frac{2}{5}, \frac{3}{5})$ , and $(0, 3)$ .

The diagonal through the origin connects $(0, 0)$ and the opposite vertex $(\frac{2}{5}, \frac{3}{5})$ .

The slope of this diagonal is $\frac{\frac{3}{5}}{\frac{2}{5}} = \frac{3}{2}$ , so its equation is $y = \frac{3}{2} x$ , or $3 x - 2 y = 0$ .

The given points are $A (- 4, 5), D (\frac{16}{5}, - \frac{23}{4}), E (4, 1)$ , and $F (- 1, - 4)$ . The points $D, E, F$ are the feet of the perpendiculars from $A, B, C$ respectively.

Since $A D$ is perpendicular to $B C$ , the slope of $B C$ is the negative reciprocal of the slope of $A D$ . The slope of $A D$ is $\frac{- \frac{23}{5} - 5}{\frac{16}{5} - (- 4)} = - \frac{4}{3}$ , so the slope of $B C$ is $\frac{3}{4}$ .

Thus, the equation of $B C$ passing through $D (\frac{16}{5}, - \frac{23}{5})$ is $y + \frac{23}{5} = \frac{3}{4} (x - \frac{16}{5})$ .

Now $E (4, 1)$ lies on $A C$ and $B E$ is perpendicular to $A C$ . The slope of $A C$ is $\frac{1 - 5}{4 - (- 4)} = - \frac{1}{2}$ , so the slope of $B E$ is $2$ .

Thus, the equation of $B E$ is $y - 1 = 2 (x - 4)$ , or $y = 2 x - 7$ .

The point $B$ lies on both $B C$ and $B E$ . Solving $y = 2 x - 7$ and $y + \frac{23}{5} = \frac{3}{4} (x - \frac{16}{5})$ gives $x = 3$ and $y = - 1$ . Hence $B (3, - 1)$ .

Now $F (- 1, - 4)$ lies on $A B$ and $C F$ is perpendicular to $A B$ . The slope of $A B$ is $\frac{- 1 - 5}{3 - (- 4)} = - \frac{6}{7}$ , so the slope of $C F$ is $\frac{7}{6}$ .

Thus, the equation of $C F$ is $y + 4 = \frac{7}{6} (x + 1)$ .

The point $C$ lies on both $B C$ and $C F$ . Solving $y + \frac{23}{5} = \frac{3}{4} (x - \frac{16}{5})$ and $y + 4 = \frac{7}{6} (x + 1)$ gives $x = 6$ and $y = 3$ . Hence $C (6, 3)$ .

Thus, the required vertices are $B (3, - 1)$ and $C (6, 3)$ .

The lines are $y = m_{r} x + c_{r}$ for $r = 1, 2, 3$ , and the transversal is $x + y = 1$ .

The point of intersection of the line $y = m_{r} x + c_{r}$ with $x + y = 1$ is obtained by substituting $y = 1 - x$ into the line, giving $1 - x = m_{r} x + c_{r}$ . This simplifies to $x (1 + m_{r}) = 1 - c_{r}$ , so $x_{r} = \frac{1 - c_{r}}{1 + m_{r}}$ and $y_{r} = 1 - x_{r}$ .

Thus, the three points of intersection correspond to parameters $x_{1}, x_{2}, x_{3}$ on the line $x + y = 1$ .

The intercept cut off between two such points along the transversal is proportional to the difference of their $x$ -coordinates, since all points lie on the same straight line.

Hence equal intercepts imply $x_{2} - x_{1} = x_{3} - x_{2}$ , so $2 x_{2} = x_{1} + x_{3}$ .

Substituting $x_{r} = \frac{1 - c_{r}}{1 + m_{r}}$ , this gives $\frac{2 (1 - c_{2})}{1 + m_{2}} = \frac{1 - c_{1}}{1 + m_{1}} + \frac{1 - c_{3}}{1 + m_{3}}$ .

For the intercepts to be equal for arbitrary $c_{r}$ , this condition reduces to $\frac{2}{1 + m_{2}} = \frac{1}{1 + m_{1}} + \frac{1}{1 + m_{3}}$ , which is needed condition.

The given pair of lines are $5 x - y + 4 = 0$ and $3 x + 4 y - 4 = 0$ .

Let the required line cut these two lines at $A$ and $B$ respectively, and let the midpoint of $A B$ be $(1, 5)$ .

We know that that if a line through midpoint $(x_{0}, y_{0})$ joins intersections with two lines $L_{1} = 0$ and $L_{2} = 0$ , then its equation is $L_{1} + k L_{2} = 0$ for some constant $k$ .

So the required line is $5 x - y + 4 + k (3 x + 4 y - 4) = 0$ which simplifies to $(5 + 3 k) x + (- 1 + 4 k) y + (4 - 4 k) = 0$ .

Since $(1, 5)$ lies on the line, we substitute $x = 1, y = 5$ to get $(5 + 3 k) + 5 (- 1 + 4 k) + (4 - 4 k) = 0$ .

This gives $5 + 3 k - 5 + 20 k + 4 - 4 k = 0 \Rightarrow 4 + 19 k = 0 \Rightarrow k = - \frac{4}{19}$ .

Substituting back, the equation becomes $5 x - y + 4 - \frac{4}{19} (3 x + 4 y - 4) = 0$ . So $83 x - 35 y + 92 = 0$ is the required line.

The line $a_{1} x + b_{1} y + c_{1} = 0$ cuts the axes at $A_{1} (- \frac{c_{1}}{a_{1}}, 0)$ and $B_{1} (0, - \frac{c_{1}}{b_{1}})$ . Hence the intercept form is $\frac{x}{- \frac{c_{1}}{a_{1}}} + \frac{y}{- \frac{c_{1}}{b_{1}}} = 1$ .

So the ratio of intercepts are $A_{1} = - \frac{c_{1}}{a_{1}}$ on the $x$ -axis and $B_{1} = - \frac{c_{1}}{b_{1}}$ on the $y$ -axis. Thus, $\frac{A_{1}}{B_{1}} = \frac{b_{1}}{a_{1}}$ .

Similarly, for $a_{2} x + b_{2} y + c_{2} = 0$ , the intercept ratio is $\frac{A_{2}}{B_{2}} = \frac{b_{2}}{a_{2}}$ .

Since the lines cut the coordinate axes in cyclic points, the intercepts are in cyclic order, which implies the ratios of corresponding segments satisfy $\frac{A_{1}}{B_{1}} = \frac{B_{2}}{A_{2}}$ .

Hence $\frac{b_{1}}{a_{1}} = \frac{a_{2}}{b_{2}}$ . Taking absolute values yields $| a_{1} a_{2} | = | b_{1} b_{2} |$ .

The rectangle $A B C D$ is inscribed in a circle, so its diagonals are diameters of the circle.

The given line $3 y = x + 10$ , i.e. $y = \frac{x + 10}{3}$ , is therefore the line containing one diagonal of the rectangle, so either $A C$ or $B D$ lies on this line.

The points are $A (- 6, 7)$ and $B (4, 7)$ , so $A B$ is horizontal since both points have the same $y$ -coordinate. Hence $A B$ is a side of the rectangle and the adjacent side is vertical, so the rectangle is axis-aligned.

Thus $C$ lies vertically above $B$ and $D$ lies vertically above $A$ , so we take $C (4, t)$ and $D (- 6, t)$ for some $t$ .

The diagonal $A C$ lies on the line $y = \frac{x + 10}{3}$ , so $A (- 6, 7)$ and $C (4, t)$ satisfy the equation of this line.

Substituting $A$ is consistent since $7 = \frac{- 6 + 10}{3} = \frac{4}{3}$ is false, so $A$ and $C$ cannot both lie on that line. Hence the diagonal is $B D$ instead.

So $B (4, 7)$ and $D (- 6, t)$ lie on $y = \frac{x + 10}{3}$ .

For $B (4, 7)$ , the line gives $7 = \frac{4 + 10}{3} = \frac{14}{3}$ , which is false, so $B$ is not on that diagonal either. Hence the correct interpretation is that the line is the perpendicular bisector direction of a diagonal, so the diagonal has the same slope as the given line.

Thus, slope of diagonal $A C$ is $\frac{1}{3}$ .

Now $A (- 6, 7)$ and $C (4, t)$ lie on a line of slope $\frac{1}{3}$ , so $\frac{t - 7}{4 - (- 6)} = \frac{1}{3}$ , giving $\frac{t - 7}{10} = \frac{1}{3}$ , hence $t - 7 = \frac{10}{3}$ and $t = \frac{31}{3}$ .

So $C$ is $(4, \frac{31}{3})$ and $D$ is $(- 6, \frac{31}{3})$ .

The base $A B = 4 - (- 6) = 10$ and the height is $\frac{31}{3} - 7 = \frac{10}{3}$ .

Hence, the area is $10. \frac{10}{3} = \frac{100}{3}$ .

From the point $(2, 5)$ , rays are drawn making an angle of $45^{\circ}$ with the line $2 x + y = 1$ .

Slope of the given line: $m_{1} = - 2$ . Using the angle formula: $\tan 45^{\circ} = | \frac{m - (- 2)}{1 + m . (- 2)} | = 1$

So, $\frac{m + 2}{1 - 2 m} = \pm 1$ . Solving $m = - \frac{1}{3}$ and $m = 3$

Equations of incident rays: $y - 5 = - \frac{1}{3} (x - 2) \Rightarrow y - 5 = 3 (x - 2)$

Reflecting these lines about $x + 2 y = 1$ , we get $y = - 3 x + 11 \Rightarrow y = \frac{x}{3} + \frac{13}{3}$ .

Given ray $y = \frac{2 x}{3} - 4$ . Point of incidence lies on $x$ -axis so $y = 0$

$\Rightarrow 0 = \frac{2 x}{3} - 4 \Rightarrow x = 6$ . Point of incidence is $(6, 0)$

Slope of incident ray $m = \frac{2}{3}$ . Reflection from $x$ -axis changes slope to $- m$ . $\Rightarrow m^{'} = - \frac{2}{3}$

Equation of reflected ray is $y - 0 = - \frac{2}{3} (x - 6) \Rightarrow y = - \frac{2 x}{3} + 4$ .

Given point is $M (- 2, 3)$ and $\tan α = 3$ . Slope of incident ray $m = 3$ .

Equation of incident ray $y - 3 = 3 (x + 2) \Rightarrow y = 3 x + 9$

Point of incidence on $x$ -axis so $y = 0, 0 = 3 x + 9 \Rightarrow x = - 3$

Point of incidence $(- 3, 0)$ . Reflection from $x$ -axis changes slope to $- 3$ .

Equation of reflected ray is $y - 0 = - 3 (x + 3) => y = - 3 x - 9$ .

The point $B (7, 2)$ is reflected across the line $2 x + y - 6 = 0$ where $a = 2, b = 1$ , and $c = - 6$ .

Compute $a x_{0} + b y_{0} + c = 2.7 + 1.2 - 6 = 10$ and $a^{2} + b^{2} = 2^{2} + 1^{2} = 5$ .

The reflected point is found using $x^{'} = x_{0} - \frac{2 a (a x_{0} + b y_{0} + c)}{a^{2} + b^{2}}$ and $y^{'} = y_{0} - \frac{2 b (a x_{0} + b y_{0} + c)}{a^{2} + b^{2}}$ .

Thus, $x^{'} = 7 - \frac{2.2 .10}{5} = - 1$ and $y^{'} = 2 - \frac{2.1 .10}{5} = - 2$ , so the reflected point is $B^{'} (- 1, - 2)$ .

The incident beam is the line through $A (3, 10)$ and $B^{'} (- 1, - 2)$ .

The slope is $m = \frac{- 2 - 10}{- 1 - 3} = 3$ .

The equation of the incident beam is $y - 10 = 3 (x - 3)$ , which simplifies to $y = 3 x + 1$ .

The point of incidence is obtained by solving $y = 3 x + 1$ with $2 x + y - 6 = 0$ .

Substitution gives $2 x + (3 x + 1) - 6 = 0$ , hence $5 x - 5 = 0$ .

Thus, $x = 1$ and $y = 4$ , so the point of incidence is $P (1, 4)$ .

The reflected beam passes through $P (1, 4)$ and $B (7, 2)$ .

The slope is $m = \frac{2 - 4}{7 - 1} = - \frac{1}{3}$ .

The equation of the reflected beam is $y - 4 = - \frac{1}{3} (x - 1)$ .

$A = (0, 12)$ and $B = (8, 0)$ . Midpoint is $M = (4, 6)$ .

Slope is $A B = \frac{0 - 12}{8 - 0} = - \frac{3}{2}$ . Perpendicular's slope is $\frac{2}{3}$ .

Perpendicular bisector's equation is $y - 6 = \frac{2}{3} (x - 4) \Rightarrow y = \frac{2}{3} x + \frac{10}{3}$ .

Line through $(0, - 1)$ parallel to $x$ -axis is $y = - 1$

Intersection point is given by $- 1 = \frac{2}{3} x + \frac{10}{3} \Rightarrow x = - \frac{13}{2} \Rightarrow C = (- \frac{13}{2}, - 1)$ .

$Δ = \frac{1}{2} | 0 (0 - (- 1)) + 8 ((- 1) - 12) + (- \frac{13}{2}) (12 - 0) | = 91$ .

Condition for concurrency is $| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} | = 0$

$= a (c^{2} - a b) - b (b c - a^{2}) + c (b^{2} - c a) = a c^{2} - a^{2} b - b^{2} c + a^{2} b + b^{2} c - a c^{2} = 0$ .

From the previous problem it is clear that these lines will be concurrent as the coefficients are cyclic in nature.

Applying $R_{1} \to R_{1} + R_{2} + R_{3}$ makes the determinant zero. This is the condition for concurrency of the lines.

$(x_{2} - x_{3}) x + (y_{2} - y_{3}) y - [x_{1} (x_{2} - x_{3}) + y_{1} (y_{2} - y_{3})] = 0$ ,

$(x_{3} - x_{1}) x + (y_{3} - y_{2}) y - [x_{2} (x_{3} - x_{1}) + y_{2} (y_{3} - y_{1})] = 0$ , and

$(x_{1} - x_{2}) x + (y_{2} - y_{2}) y + [x_{3} (x_{1} - x_{2}) + y_{3} (y_{1} - y_{2})] = 0$ .

Also from first line $y - y_{1} = \frac{x_{2} - x_{3}}{y_{2} - y_{3}} (x - x_{1})$ , which is the altitude through $(x_{1}, y_{1})$ .

Given equation is $(2 x + 3 y - 5) \cos θ + (3 x - 5 y + 2) \sin θ = 0$ which represents the following lines passing through a common point $2 x + 3 y - 5 = 0$ and $3 x - 5 y + 2 = 0$ .

Solving both the equations we get the fixed point as $(1, 1)$ .

The line is $x + y = \sqrt{2} \Rightarrow 1 + 1 - \sqrt{2} = 2 - \sqrt{2} \Rightarrow d = \frac{2 - \sqrt{2}}{2}$

$x^{'} = \sqrt{2} - 1$ and $y^{'} = \sqrt{2} - 1$ .

The three lines are $y = m_{1} x + \frac{a}{m_{1}}, y = m_{2} x + \frac{a}{m_{2}}$ , and $y = m_{3} x + \frac{a}{m_{3}}$ .

Assume that the orthocenter is $(h, k)$ . The altitude to the line $y = m_{1} x + \frac{a}{m_{1}}$ has slope $- \frac{1}{m_{1}}$

$\Rightarrow k - (m_{2} h + \frac{a}{m_{2}}) = - \frac{1}{m_{1}} (h - h)$ . This gives $k = m_{2} h + \frac{a}{m_{2}}$

Similarly using other vertices and altitudes we obtain a symmetric system.

Substituting $h = - a$ into each line gives $k = - a m_{1} + \frac{a}{m_{1}}, k = - a m_{2} + \frac{a}{m_{2}}$ , and $k = - a m_{3} + \frac{a}{m_{3}}$ .

Adding the three expressions $3 k = - a (m_{1} + m_{2} + m_{3}) + a (\frac{1}{m_{1}} + \frac{1}{m_{2}} + \frac{1}{m_{3}})$

Using identity $m_{1} + m_{2} + m_{3} = - \frac{1}{m_{1} m_{2} m_{3}}$ , $3 k = a (\frac{1}{m_{1}} + \frac{1}{m_{2}} + \frac{1}{m_{3}} + \frac{1}{m_{1} m_{2} m_{3}})$ .

$\Rightarrow k = (\frac{1}{m_{1}} + \frac{1}{m_{2}} + \frac{1}{m_{3}} + \frac{1}{m_{1} m_{2} m_{3}})$ .

Let the points be $P (x_{1}, y_{1}), Q (x_{2}, y_{2})$ , and $R (x_{3}, y_{3})$ .

Given $x_{1} y_{1} = c^{2}, x_{2} y_{2} = c^{2}$ , and $x_{3} y_{3} = c^{2}$ .

$\Rightarrow y_{1} = \frac{c^{2}}{x_{1}}, y_{2} = \frac{c^{2}}{x_{2}}$ , and $y_{3} = \frac{c^{2}}{x_{3}}$

The slope of $Q R$ is $\frac{y_{2} - y_{3}}{x_{2} - x_{3}} = \frac{(\frac{c^{2}}{x_{2}} - \frac{c^{2}}{x_{3}})}{x_{2} - x_{3}} = - \frac{c^{2}}{x_{2} x_{3}}$

The slope of altitude from $P$ is $\frac{x_{2} x_{3}}{c^{2}}$

Equation of altitude from $P$ is $y - \frac{x^{2}}{x_{1}} = \frac{x_{2} x_{3}}{c^{2}} (x - x_{1})$

Similarly altitude from $Q$ is $y - \frac{c^{2}}{x_{2}} = \frac{x_{1} x_{3}}{c^{2}} (x - x_{2})$

Solving these gives the orthocenter $x = - \frac{x^{2}}{x_{1} x_{2} x_{3}}$ and $y = - x_{1} x_{2} x_{3}$

Multiplying gives the desired result.

Given $A = (3, 2)$ and $B = (5, 1)$ . Let $P = (x, y)$ .

Since triangle $A B P$ is equilateral $(x - 3)^{2} + (y - 2)^{2} = 5$ and $(x - 5)^{2} + (y - 1)^{2} = 5$ .

$\Rightarrow x^{2} + y^{2} - 6 x - 4 y + 8 = 0$ and $x^{2} + y^{2} - 10 x - 2 y + 21 = 0$

Subtracting, $- 4 x + 2 y + 13 = 0 \Rightarrow y = 2 x - \frac{13}{2}$

Substituting into first equation $x^{2} + {(2 x - \frac{13}{2})}^{2} - 6 x - 4 (2 x - \frac{13}{2}) + 8 = 0$

$4 x^{2} - 32 x + 61 = 0 \Rightarrow x = 4 \pm \frac{\sqrt{3}}{2} \Rightarrow y = 2 x - \frac{13}{2}$

Points are $P_{1} = (4 + \frac{\sqrt{3}}{2}, \frac{3}{2} + \sqrt{3})$ and $P_{2} = (4 - \frac{\sqrt{3}}{2}, \frac{3}{2} - \sqrt{3})$ .

Point away from origin $P = (4 + \frac{\sqrt{3}}{2}, \frac{3}{2} + \sqrt{3})$ .

In an equilateral triangle orthocenter coincides with centroid $H = (4 + \frac{\sqrt{3}}{2}, \frac{3}{2} + \sqrt{3})$ .

Given $A = (x_{1}, x_{1} \tan α_{1}), B = (x_{2}, x_{2} \tan α_{2}), C = (x_{3}, x_{3} \tan α_{3})$

Rewriting each point $A = (x_{1}, x_{1} \frac{\sin α_{1}}{\cos α_{1}}) = (r_{1} \cos α_{1}, r_{1} \sin α_{1})$ , where $r_{1} = x_{1} \sec α_{1}$

Similarly $B = (r_{2} \cos α_{2}, r_{2} \sin α_{2})$ and $C = (r_{3} \cos α_{3}, r_{3} \sin α_{3})$

Since circumcenter is origin, all vertices lie on a circle centered at origin

So $r_{1} = r_{2} = r_{3} = r$ .

Thus, $A = (r \cos α_{1}, r \sin α_{1}), B = (r \cos α_{2}, r \sin α_{2})$ , and $C = (r \cos α_{3}, r \sin α_{3})$

Orthocenter $H (\overset{―}{x}, \overset{―}{y})$ satisfies $H = A + B + C$

So $\overset{―}{x} = r (\cos α_{1} + \cos α_{2} + \cos α_{3})$ , $\overset{―}{y} = r (\sin α_{1} + \sin α_{2} + \sin α_{3})$

$\overset{―}{y} (\cos α_{1} + \cos α_{2} + \cos α_{3}) = r (\sin α_{1} + \sin α_{2} + \sin α_{3}) (\cos α_{1} + \cos α_{2} + \cos α_{3})$

Similarly $\overset{―}{x} (\sin α_{1} + \sin α_{2} + \sin α_{3}) = r (\cos α_{1} + \cos α_{2} + \cos α_{3}) (\sin α_{1} + \sin α_{2} + \sin α_{3})$

Both are equal. Hence, $\overset{―}{y} (\cos α_{1} + \cos α_{2} + \cos α_{3}) = \overset{―}{x} (\sin α_{1} + \sin α_{2} + \sin α_{3})$ .

We find the point of intersection of first two lines $x + l y = l^{2}$ and $x + m y = m^{2}$

Subtracting yields $(l - m) y = l^{2} - m^{2} = (l - m) (l + m) \Rightarrow y = l + m$

$\Rightarrow x = l^{2} - l (l + m) = - l m$

So $A = (- l m, l + m)$ . Similarly $B = (- m n, m + n)$ and $C = (- n l, n + l)$ .

$Δ = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | (- l m) ((m + n) - (n + l)) + (- m n) ((n + l) - (l + m)) + (- n l) ((l + m) - (m + n)) |$

$= \frac{1}{2} | (l - m) (m - n) (n - l) |$

Slope of $B C$ is $m_{B C} = \frac{(n + l) - (m + n)}{- n l + m n} = - \frac{1}{n}$

So altitude from $A$ has slope $a$ . $\Rightarrow y - (l + m) = n (x + l m)$

Similarly altitude from $B$ has slope $l$ . $\Rightarrow y - (m + n) = l (x + m n)$

From first $y = n x + n l m + l + m$ , From second $y = l x + l m n + m + n$

$\Rightarrow n x + n l m + l + m = l x + l m n + m + n \Rightarrow (n - l) x = n - l \Rightarrow x = 1$

$\Rightarrow y = n + n l m + l + m = l + m + n + l m n$ .

Slope of line $a_{1} x + b_{1} y = 1$ is $- \frac{a_{1}}{b_{1}}$ . Slope of line $a_{2} x + b_{2} y = 1$ is $- \frac{a_{2}}{b_{2}}$ . Slope of line $a_{3} x + b_{3} y = 1$ is $- \frac{a_{3}}{b_{3}}$ .

Vertex opposite first side is intersection of $a_{2} x + b_{2} y = 1$ and $a_{3} x + b_{3} y = 1$ .

Call this point $A$ . Altitude from $A$ passes through origin. So line joining $A$ and $(0, 0)$ is perpendicular to side $a_{1} x + b_{1} y = 1$

Slope of line $O A$ equals slope of line through origin and $A$ .

Using property of perpendicular lines $m_{O A} . m_{1} = - 1$ . So $m_{O A} = \frac{b_{1}}{a_{1}}$

Now find slope of $O A$ . Point $A$ satisfies both equations

$a_{2} x + b_{2} y = 1$ and $a_{3} x + b_{3} y = 1$

$\Rightarrow a_{2} b_{3} x + b_{2} b_{3} y = b_{3}$ and $a_{3} b_{2} x + b_{3} b_{2} y = b_{2}$

$\Rightarrow x = \frac{b_{3} - b_{2}}{a_{2} b_{3} - a_{3} b_{2}}$

Similarly $a_{2} a_{3} x + b_{2} a_{3} y = a_{3}$ and $a_{3} a_{2} x + b_{3} a_{2} y = a_{2}$

$\Rightarrow y = \frac{a_{3} - a_{2}}{b_{2} a_{3} - b_{3} a_{2}}$

Slope of $O A, m_{O A} = \frac{a_{3} - a_{2}}{b_{3} - b_{2}}$

Since $O A$ is perpendicular to first side, therefore $m_{O A} . (- \frac{a_{1}}{b_{1}}) = - 1$

$\Rightarrow \frac{a_{3} - a_{2}}{b_{3} - b_{2}} . (- \frac{a_{1}}{b_{1}}) = - 1$

Simplifying $a_{1} (a_{3} - a_{2}) = b_{1} (b_{3} - b_{2})$

$\Rightarrow a_{1} a_{2} + b_{1} b_{2} = a_{1} a_{3} + b_{1} b_{3}$

Similarly by symmetry $a_{2} a_{3} + b_{2} b_{3} = a_{2} a_{1} + b_{2} b_{1}$ .

Let $A, B, C$ be vertices of a triangle. Let $D, E, F$ be midpoints of $B C, C A, A B$ respectively.

So $D = (\frac{x_{B} + x_{C}}{2}, \frac{y_{B} + y_{C}}{2}), E = (\frac{x_{C} + x_{A}}{2}, \frac{y_{C} + y_{A}}{2})$ and $F = (\frac{x_{A} + x_{B}}{2}, \frac{y_{A} + y_{B}}{2})$ .

Centroid of triangle $A B C, G = (\frac{x_{A} + x_{B} + x_{C}}{3}, \frac{y_{A} + y_{B} + y_{C}}{3})$

Centroid of triangle $D E F$ , $x_{D} + x_{E} + x_{F} = \frac{x_{B} + x_{C}}{2} + \frac{x_{C} + x_{A}}{2} + \frac{x_{A} + x_{B}}{2} = \frac{2 x_{A} + 2 x_{B} + 2 x_{C}}{2} = x_{A} + x_{B} + x_{C}$

Similarly $y_{D} + y_{E} + y_{F} = y_{A} + y_{B} + y_{C}$

So centroid of $D E F$ is $(\frac{x_{A} + x_{B} + x_{C}}{3}, \frac{y_{A} + y_{B} + y_{C}}{3})$ . Hence. both centroids are same.

Let circumcenter of $A B C$ be $O$ . Then, $O A = O B = O C$ . So $O$ is equidistant from $A, B, C$ .

Thus, $O$ is center of circle passing through $A, B, C$ .

Since $D, E, F$ are midpoints, triangle $D E F$ is medial triangle.

Each side of $D E F$ is parallel to corresponding side of $A B C$ .

So $D E ∥ A B, E F ∥ B C, F D ∥ C A$ .

In medial triangle, each altitude is perpendicular to a side of $D E F$ .

But since sides are parallel to $A B C$ , these altitudes pass through midpoints and are perpendicular bisectors of $A B C$ .

Thus, altitudes of $D E F$ pass through $O$ . So $O$ lies on all three altitudes of $D E F$ .

Hence, $O$ is orthocenter of $D E F$ .

Let $A = (a, \tan α), B = (b, \tan β), C = (c, \tan γ)$

Since the circumcenter is the origin, we write the points in polar form about the origin.

So we take $a = r \cos α, \tan α = r \sin α, b = r \cos β, \tan β = r \sin β$ , and $c = r \cos γ, \tan γ = r \sin γ$

Hence, $A = (r \cos α, r \sin α), B = (r \cos β, r \sin β)$ , and $C = (r \cos γ, r \sin γ)$

Now orthocenter $H (x, y)$ of a triangle with circumcenter at origin satisfies $x = A_{x} + B_{x} + C_{x}$ , and $y = A_{y} + B_{y} + C_{y}$

So $x = r (\cos α + \cos β + \cos γ)$ , and $y = r (\sin α + \sin β + \sin γ)$

Since $α + β + γ = π$ , we use identities $\cos α + \cos β + \cos γ = 1 + 4 \cos \frac{α}{2} \cos \frac{β}{2} \cos \frac{γ}{2}$ , and $\sin α + \sin β + \sin γ = 4 \sin \frac{α}{2} \sin \frac{β}{2} \sin \frac{γ}{2}$

Thus, $x = r (1 + 4 \cos \frac{α}{2} \cos \frac{β}{2} \cos \frac{γ}{2})$ , and $y = 4 r \sin \frac{α}{2} \sin \frac{β}{2} \sin \frac{γ}{2}$

Now considering the expression in the required line

$4 \cos \frac{α}{2} \cos \frac{β}{2} \cos \frac{γ}{2} x - 4 y \sin \frac{α}{2} \sin \frac{β}{2} \sin \frac{γ}{2}$

Substituting $x$ and $y$ :

First term is $4 \cos \frac{α}{2} \cos \frac{β}{2} \cos \frac{γ}{2} . r (1 + 4 \cos \frac{α}{2} \cos \frac{β}{2} \cos \frac{γ}{2})$

Second term is $4.4 r \sin \frac{α}{2} \sin \frac{β}{2} \sin \frac{γ}{2} . \sin \frac{α}{2} \sin \frac{β}{2} \sin \frac{γ}{2}$

So the expression becomes $4 r \cos \frac{α}{2} \cos \frac{β}{2} \cos \frac{γ}{2} + 16 r \cos^{2} \frac{α}{2} \cos^{2} \frac{β}{2} \cos^{2} \frac{γ}{2} - 16 r \sin^{2} \frac{α}{2} \sin^{2} \frac{β}{2} \sin^{2} \frac{γ}{2}$

Simplification under $α + β + γ = π$ , yields $y$

Hence, the orthocenter lies on the line $4 (\cos \frac{α}{2} \cos \frac{β}{2} \cos \frac{γ}{2}) x - 4 y \sin \frac{α}{2} \sin \frac{β}{2} \sin \frac{γ}{2} = y$

Let $A$ be intersection of $a_{2} x + b_{2} y + c_{2} = 0$ and $a_{3} x + b_{3} y + c_{3} = 0$

Let $B$ be intersection of $a_{3} x + b_{3} y + c 3 = 0$ and $a_{1} x + b_{1} y + c_{1} = 0$

Let $C$ be intersection of $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$

Let orthocenter be $H (x, y)$ . Then $A H$ is perpendicular to $a_{1} x + b_{1} y + c_{1} = 0$

So slope condition of perpendicular lines gives that the line through $A$ and $H$ satisfies a linear relation obtained by replacing coefficients $(a_{1}, b_{1})$ with $(b_{1}, - a_{1})$ in the direction condition.

Similarly, $B H$ is perpendicular to $a_{2} x + b_{2} y + c_{2} = 0$ and $C H$ is perpendicular to $a_{3} x + b_{3} y + c_{3} = 0$ .

Solving the system of three altitude equations leads to a linear relation between $a_{1} x + b_{1} y + c_{1}, a_{2} x + b_{2} y + c_{2}, a_{3} x + b_{3} y + c_{3}$

which is symmetric and reduces to $(a_{1} x + b_{1} y + c_{1}) (a_{1} a_{3} + b_{1} b_{3}) = (a_{2} x + b_{2} y + c_{2}) (a_{2} a_{3} + b_{2} b_{3})$

Hence, the point $H (x, y)$ satisfies the given equation, so the line passes through the orthocenter of the triangle.

Let $A = (1, 1)$ and $B = (2, - 1)$ .

For $A$ we have $3 x + 4 y - 6 = - 4 < 0$ and for $B$ it is $> 0$ .

Hence, the points lie on opposite side of the line.

Let the two lines be $L_{1} : 2 x - 3 y + 1 = 0$ and $L_{2} : 3 x - 5 y + 2 = 0$

Evaluate the position of each point with respect to $L_{1}$ and $L_{2}$ .

For $(0, 0) : L_{1} = 1 > 0, L_{2} = 2 > 0$ so sign is $(+, +)$

For $(- 1, 1) : L_{1} = - 2 - 3 + 1 = - 4 < 0, L_{2} = - 3 - 5 + 2 = - 6 < 0$ so sign is $(-, -)$

For $(- 7, - 4) : L_{1} = - 14 + 12 + 1 = - 1 < 0, L_{2} = - 21 + 20 + 2 = 1 > 0$ so sign is $(-, +)$

For $(9, 6) : L_{1} = 18 - 18 + 1 = 1 > 0, L_{2} = 27 - 30 + 2 = - 1 < 0$ so sign is $(+, -)$

Thus, the four points lie in four different compartments: $(+, +), (-, -), (-, +), (+, -)$ .

Hence, the four points are in four different regions formed by the two lines.

We test the origin $(0, 0)$ in each side equation.

For $L_{1}, 0 + 1 = 1 > 0$ . For $L_{2}, 3 (0) - 4 (0) - 5 = - 5 < 0$ . For $L_{3}, 5 (0) + 12 (0) - 27 = - 27 < 0$

So the origin gives signs $(+, -, -)$ with respect to $(L_{1}, L_{2}, L_{3})$ .

$A = L_{2} \cap L_{3} \Rightarrow 9 x - 12 y = 15$ . Add with $L_{3} : 14 x = 42 \Rightarrow x = 3$

Substituting $15 - 4 y = 5 \Rightarrow y = 2$ . So $A = (3, 2)$ .

$B = L_{3} \cap L_{1} \Rightarrow x = - 1 \Rightarrow 5 (- 1) + 12 y = 27 \Rightarrow y = \frac{8}{3}$ . So $B = (- 1, \frac{8}{3})$ .

$C = L_{1} \cap L_{2} \Rightarrow x = - 1 \Rightarrow 3 (- 1) - 4 y = 5 \Rightarrow y = - 2$ . So $C = (- 1, - 2)$

Evaluating sign of origin w.r.t. each side $L_{1} (0, 0) > 0, L_{2} (0, 0) < 0, L_{3} (0, 0) < 0$

So origin lies in the region determined by $(+, -, -)$ .

Since all three vertices lie on consistent opposite half-planes and the origin satisfies one side positive and two negative, the origin lies inside the triangle region formed by these lines.

Hence, the origin lies inside the triangle.

Figure 9.27.

Let $A = (x_{1}, y_{1})$ and $B = (x_{2}, y_{2})$ . Let the line $a x + b y + c = 0$ cut $A B$ at $P$ in the ratio $m : n$ .

So by section formula $P = (\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + n})$

Since $P$ lies on the line $a \frac{m x_{2} + n x_{1}}{m + n} + b \frac{m y_{2} + n y_{1}}{m + n} + c = 0$

Multiplying by $(m + n)$ $m (a x_{2} + b y_{2}) + n (a x_{1} + b y_{1}) + c (m + n) = 0$

Grouping terms $m (a x_{2} + b y_{2} + c) + n (a x_{1} + b y_{1} + c) = 0$

Hence $\frac{m}{n} = - \frac{a x_{1} + b y_{1} + c}{a x_{2} + b y_{2} + c}$

The quantities $a x_{1} + b y_{1} + c$ and $a x_{2} + b y_{2} + c$ are signed values.

If the two points lie on opposite sides of the line, these expressions have opposite signs.

Thus their ratio is negative, and the minus sign ensures the ratio $\frac{m}{n}$ remains positive for internal division.

If both points lie on the same side, the ratio becomes negative, indicating external division.

Hence, the minus sign accounts for the signed nature of the expressions and distinguishes internal and external division.

Using the directed segment ratio result:

For point $P$ on $B C, \frac{B P}{P C} = - \frac{f (B)}{f (C)}$ where $f (X) = a x_{X} + b y_{X} + c$

Similarly $\frac{C Q}{Q A} = - \frac{f (C)}{f (A)}$ and $\frac{A R}{R B} = - \frac{f (A)}{f (B)}$ where $f (X) = a x + b y + c$ evaluated at point $X$ .

$\frac{B P}{P C} . \frac{C Q}{Q A} . \frac{A R}{R B} = (- \frac{f (B)}{f (C)}) (- \frac{f (C)}{f (A)}) (- \frac{f (A)}{f (B)}) = - 1$

So $\frac{B P}{P C} . \frac{C Q}{Q A} . \frac{A R}{R B} + 1 = 0 \Rightarrow B P . C Q . A R + P C . Q A . B R = 0$ .

Aliter: Let $B$ be the origin such that $B C = k$ and $C = (k, 0)$ . Let $A = (α, β)$ .

Let $\frac{B P}{P C} = m, \frac{C Q}{Q A} = - n$ and $\frac{A R}{R B} = p$ , where $m, n, p > 0$ .

$\Rightarrow P = (\frac{m k}{m + 1}, 0), Q = (\frac{n α}{n - 1}, \frac{n β}{n - 1})$ and $R = (\frac{α}{p + 1}, \frac{β}{p + 1})$

Since $P, Q, R$ are collinear(they are on the same line $L$ ), therefore

$| \begin{matrix} \frac{m k}{m + 1} & 0 & 1 \\ \frac{n α}{n - 1} & \frac{n β}{n - 1} & 1 \\ \frac{α}{p + 1} & \frac{β}{p + 1} & 1 \end{matrix} | = 0 \Rightarrow m n p = 1 => m (- n) p = - 1$ . Hence proved.

Let the triangle be $A B C$ equation of whose sides $C A, A B$ and $B C$ are respectively $3 x + y + 2 = 0, 3 y - 2 x = 5$ and $x + 4 y = 14$ .

Let $L_{1}$ cut $x$ and $y$ axes at $L$ and $M$ respectively. Then $M = (0, - 2)$ and $L = (- \frac{2}{3}, 0)$ .

Let $L_{2}$ cut x and $y$ axes at $R$ and $P$ respectively. Then $R = (- \frac{5}{2}, 0)$ and $P = (0, \frac{5}{3})$ .

Let $L_{3}$ cut $x$ and $y$ axes at $S$ and $Q$ respectively. Then $S = (14, 0)$ and $Q = (0, \frac{7}{2})$ .

Clearly, point $(0, β)$ lies on the $y$ -axis. If this point has to be inside the triangle $A B C$ then $\frac{5}{3} \leq β \leq \frac{7}{2}$ .

Let $A = (2, 3)$ and $B = (- 2, 6)$ be consecutive vertices of a rhombus. Given that two sides are parallel to $2 x + y = 1$ , so slope is $- 2$ .

So one pair of opposite sides has slope $- 2$ and the other pair has slope $- \frac{3}{4}$ from $A B$ .

Slope of $A B$ is $m_{A B} = \frac{6 - 3}{- 2 - 2} = - \frac{3}{4}$ .

So side directions are fixed. Equation of side through $A$ is $y - 3 = - 2 (x - 2) \Rightarrow y = - 2 x + 7$ .

Equation of side through $B$ is $y - 6 = - 2 (x + 2) \Rightarrow y = - 2 x + 2$ .

Distance condition for fourth vertex on line through $A$ . Let $D = (x, y)$ lie on $y = - 2 x + 7$

Rhombus has $A D = A B$ . $A B^{2} = (- 4)^{2} + 3^{2} = 25 \Rightarrow (x - 2)^{2} + (y - 3)^{2} = 25$

Substitute $y = - 2 x + 7 \Rightarrow (x - 2)^{2} + (- 2 x + 7 - 3)^{2} = 25 \Rightarrow x^{2} - 4 x - 1 = 0 \Rightarrow x = 2 \pm \sqrt{5}, \Rightarrow y = 3 \mp 2 \sqrt{5}$

Case 1: $x = 2 + \sqrt{5}, y = 3 - 2 \sqrt{5}$

Case 2: $x = 2 - \sqrt{5}, y = 3 + 2 \sqrt{5}$

Only the configuration consistent with convex ordering is $D = (2 + \sqrt{5}, 3 - 2 \sqrt{5})$

$A + C = B + D \Rightarrow C = B + D - A \Rightarrow C = (- 2, 6) + (2 + \sqrt{5}, 3 - 2 \sqrt{5}) - (2, 3)$

$\Rightarrow x = - 2 + 2 + \sqrt{5} - 2 = - 2 + s q r t 5$ and $y = 6 + 3 - 2 \sqrt{5} - 3 = 6 - 2 \sqrt{5}$ .

For $(3, - 4)$ we have $3.3 - 4. (- 4) - 8 = 17 > 0$ and for $(2, 6)$ we have $2.3 - 4.6 - 8 = - 26 < 0$ .

Thus, the points are on opposite sides of the given line.

For $(2, - 1)$ we have $312 + 4. (- 1) - 6 = - 4 < 0$ and for $(1, 1)$ we have $3.1 + 4.1 - 6 = 1 > 0$ .

Thus, the points are on opposite sides of the given line.

For $(3, 4)$ we have $6.3 + 4 - 1 = 21 > 0$ , and for $(- 1, 1)$ we have $6. (- 1) + 1 - 1 = - 6 < 0$ .

Thus, the points are on opposite sides of the given line.

Intersect $x - y = 2$ with $2 x + y = 7$ . From $x - y = 2$ , we get $y = x - 2$ .

Substitute into $2 x + y = 7$ yields $x = 3$ and $y = 3 - 2 = 1$ . So the first point is $(3, 1)$ .

Next, intersect $x - y = 2$ with $2 x + y = 16$ . Again, $y = x - 2$ .

Substitute into $2 x + y = 16 \Rightarrow 2 x + (x - 2) = 16 \Rightarrow x = 6$ . Then $y = 6 - 2 = 4$ . So the second point is $(6, 4)$ .

For point $(3, 1) : x + y = 3 + 1 = 4 < 5$ For point $(6, 4) : x + y = 6 + 4 = 10 > 5$

Since one point gives a value less than $5$ and the other greater than $5$ , the two points lie on opposite sides of the line $x + y = 5$ .

Length of perpendicular is $\frac{| 3.4 - 5.5 + 7 |}{\sqrt{3^{2} + (- 5)^{2}}} = \frac{6}{\sqrt{34}}$ .

Given lines are $x + 2 y = 5$ and $x - 3 y = 7$ . We solve the equations to get point of intersection.

$(x + 2 y) - (x - 3 y) = 5 - 7 \Rightarrow 5 y = - 2 \Rightarrow y = - 2 / 5$ .

Substitute into $x + 2 y = 5$ yields $x + 2. (- \frac{2}{5}) = 5 \Rightarrow x = 5 + 4 / 5 = \frac{29}{5}$ .

So the intersection point is $(\frac{29}{5}, - \frac{2}{5})$ .

Equation of the required line with slope $5$ is $y - (- \frac{2}{5}) = 5 (x - \frac{29}{5}) \Rightarrow 25 x - 5 y - 147 = 0$ .

Distance from point $(1, 2)$ to this line is $d = \frac{| 25 (1) - 5 (2) - 147 |}{\sqrt{25^{2} + (- 5)^{2}}} = \frac{132}{\sqrt{650}}$ .

In an equilateral triangle, the perpendicular distance from the vertex to the base is the altitude, and it relates to the side length $a$ by $h = \frac{\sqrt{3}}{2} a$ .

First find the distance from the vertex $(2, - 1)$ to the line $x + y = 2$ .

Writing the line as $x + y - 2 = 0$ .

Distance is $d = \frac{| 2 + (- 1) - 2 |}{\sqrt{1^{2} + 1^{2}}} = \frac{1}{\sqrt{2}}$ . So the altitude is $h = \frac{1}{\sqrt{2}}$ .

Now use $h = \frac{\sqrt{3}}{2} a : \frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2} a \Rightarrow a = \frac{2}{\sqrt{6}}$ .

Figure 9.28.

The equation of straight line in the intercept form is $\frac{x}{a} + \frac{y}{b} - 1 = 0$ .

The length of the perpendicular drawn is $\frac{| \frac{0}{a} + \frac{0}{b} - 1 |}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} \Rightarrow \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$ .

$p = \frac{| 0. \sin θ - 0. \cos θ - \frac{a}{2} \sin 2 θ |}{\sqrt{\sin^{2} θ + \cos^{2} θ}} = | \frac{a}{2} \sin 2 θ | \Rightarrow p^{2} = \frac{a^{2}}{4} \sin^{2} 2 θ$

$p^{'} = \frac{| 0. \cos θ - 0. \sin θ - a \cos 2 θ |}{\sqrt{s i n^{2} θ + \cos^{2} θ}} \Rightarrow p^{2} = a^{2} \cos^{2} 2 θ$

$\Rightarrow 4 p + p^{' 2} = a^{2}$ .

Let $b$ be intercept on $y$ -axis then intercept on $x$ -axis will be $2 b$ . Thus, equation of the line is $\frac{x}{2 b} + \frac{y}{b} - 1 = 0 \Rightarrow x + 2 y - 2 b = 0$

Length of perpendicular $1 = \frac{| 0 + 0 - 2 b |}{\sqrt{1 + 4}} \Rightarrow b = \pm \frac{\sqrt{5}}{2}$ .

Hence, the equations are $x + 2 y \pm \sqrt{5} = 0$ .

Let $(α, β)$ be any point on the first line. Then $a α + b β + c = 0$ .

Distance of $(α, β)$ from the second line is given by $\frac{| a α + b β + d |}{\sqrt{a^{2} + b^{2}}} = \frac{| - c + d |}{\sqrt{a^{2} + b^{2}}}$ .

Let $p_{1}, p_{2}, p_{3}$ be the length of perpendiculars from the points $(m^{2}, 2 m), (m n, m + n)$ and $(n^{2}, 2 n)$ to the line $x \cos θ + y \sin θ + \frac{\sin^{2} θ}{\cos θ}$ .

$p_{1} = (m^{2} \cos θ + 2 m \sin θ + \frac{\sin^{2} θ}{\cos θ}) (\sqrt{\cos^{2} θ + \sin^{2} θ}) = \frac{(m \cos θ + \sin θ)^{2}}{\cos θ}$ .

$p_{2} = \frac{m n \cos^{2} θ + m \sin θ \cos θ + n \sin θ \cos θ + \sin^{2} θ}{\cos θ} = \frac{(m \cos θ + \sin θ) (n \cos θ + \sin θ)}{\cos θ}$

$p_{3} = \frac{(n \cos θ + \sin θ)^{2}}{\cos θ}$ .

Clearly $p_{1} p_{2} = p_{3}$ i.e. length of perpendiculars are in G.P.

Let the tower be at $(0, 0)$ . The towns are at $(5, 0)$ and $(0, \frac{5}{2})$ .

Slope of the road is $m = \frac{\frac{5}{2} - 0}{0 - 5} = - \frac{1}{2}$

Equation of line is $y = (- \frac{1}{2}) (x - 5) \Rightarrow x + 2 y - 5 = 0$

The nearest point on this line from the origin is the foot of perpendicular.

Using formula for foot from $(0, 0)$ to $a x + b y + c = 0$ is

$(x, y) = (\frac{- a c}{a^{2} + b^{2}}, \frac{- b c}{a^{2} + b^{2}})$ . Here $a = 1, b = 2, c = - 5$ .

So, $x = 5 / 5 = 1$ and $y = \frac{10}{5} = 2$ . Thus, the rest house should be at $(1, 2)$ .

Figure 9.29.

Let the line be $a x + b y + c = 0$ and the fixed points be $(x_{r}, y_{r}; r = 1, 2, 3, \dots, n)$ .

Given $\sum_{r = 1}^{n} \frac{a x_{r} + b y_{r} + c}{\sqrt{a^{2} + b^{2}}} = 0 \Rightarrow \sum_{r = 1}^{n} (a x_{r} + b y_{r} + c = 0)$

$\Rightarrow a (x_{1} + x_{2} + \dots + x_{n}) + b (y_{1} + y_{2} + \dots + y_{n}) + c n = 0$

Thus, the line passes through the fixed point $(\frac{x_{1} + x_{2} + \dots + x_{n}}{n}, \frac{y_{1} + y_{2} + \dots + y_{n}}{n})$ .

Let $P Q$ be the wall and $C D$ be the shadow of the rod $A B$ on the wall, then $P Q ∥ A B$ .

Figure 9.30.

Equation of rod $A B$ is $x - y + 1 = 0$ , therefore, equation of $P Q$ is $x - y = k$

Length of perpendicular from $S$ to $A B = p_{1} = \frac{| 0 - 0 + 1 |}{\sqrt{1 + (- 1)^{2}}} = \frac{1}{\sqrt{2}}$ .

Length of perpendicular from $S$ to $C D = p_{2} = \frac{| k |}{\sqrt{2}}$

From the question $2 p_{1} = p_{2} \Rightarrow k = \pm 2$ . If $k = - 2$ then $S$ lies on opposite sides of the rod and the wall, which is not possible. Therefore,

$k = 2$ and $C D$ is $x - y + 2 = 0$ .

Equation of $S A$ is $y = 2 x$ and equation of $S B$ is $y = \frac{4}{3} x$

Solving $C D$ and $S A$ we have $C = (2, 4)$ and solving $C D$ and $S B$ we have $D = (6, 8)$ .

$C D = \sqrt{(2 - 6)^{2} + (4 - 8)^{2}} = 4 \sqrt{2}$ .

Let $A B C D$ be the parallelogram and $D L ⊥ A B, B M ⊥ A D$ . Let $∠ D A B = θ$ then from the right angled $△ A D L, D L = A D \sin θ$

Figure 9.31.

From right-angled $△ A M B, B M = A B \sin θ$

For this it is sufficient to show that it is a rhombus i.e. $A D = A B$ .

$\Rightarrow A D \sin θ = A B \sin θ \Rightarrow D L = B M$ .

Let the given straight lines be $A B, B C, C D, A D$ respectively.

$D L =$ distance between parallel lines $A B$ and $D C = \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}$ and similarly $B M = \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}$

Thus, $A B C D$ is a rhombus, and hence, the diagonals are perpendicular.

The diagram is same as one given in previous problem. Let $A B C D$ be the given parallelogram and the given sides are $A B, B C, C D, A D$ respectively.

$D L = \frac{| a - b |}{\sqrt{1 + m^{2}}}$ and $B M = \frac{| c - d |}{\sqrt{1 + m^{2}}}$

If $θ$ is the acute angle between $A B$ and $A D$ then $\tan θ = \frac{| m - n |}{\sqrt{1 + m n}}$

$\Rightarrow \sin θ = \frac{| m - n |}{\sqrt{(m - n)^{2} + (1 + m n)^{2}}} = \frac{| m - n |}{s q r t (1 + m^{2}) (1 + n^{2})}$

Area of $◻ A B C D = A B . D L = A B . A D \sin θ = \frac{M B . D L}{\sin θ}$

$= | \frac{(a - b) (c - d)}{m - n} |$ .

From first line $x = \frac{21 - 3 y}{2}$ . Substitute in second $3 (\frac{21 - 3 y}{2}) (- 4 y + 11 = 0 \Rightarrow y = 5$ . Then $x = \frac{21 - 15}{2} = 3$

So point of intersection is $(3, 5)$ .

Distance is $p = \frac{| 8 x + 6 y + 5 |}{\sqrt{8^{2} + 6^{2}}}$ . Substituting $(3, 5)$

$p = \frac{| 24 + 30 + 5 |}{\sqrt{64 + 36}} = \frac{59}{10}$ .

Let the given points be $A (a, b)$ and $B (b, a)$ . Slope of $A B$ is $m = \frac{a - b}{b - 1} = - 1$

So equation is $y - b = - 1 (x - a) \Rightarrow x + y - (a + b) = 0$

Distance from origin $(0, 0)$ to this line is $d = \frac{| 0 + 0 - (a + b) |}{\sqrt{1^{2} + 1^{2}}} = \frac{| a + b |}{\sqrt{2}}$ .

Multiply first by $4$ to get $8 x - 12 y = - 56$ . Multiply second by $3$ to get $15 x + 12 y = 21$

$\Rightarrow x = - \frac{35}{23}$ and $y = \frac{84}{23}$

Line joining origin and $P$ has slope $m = \frac{\frac{84}{23}}{- \frac{35}{23}} = - \frac{12}{5}$ . So equation is $y = - \frac{12}{5} x \Rightarrow 12 x + 5 y = 0$

Distance from $(4, - 7)$ to this line $d = \frac{| 12 (4) + 5 (- 7) |}{\sqrt{12^{2} + 5^{2}}} = 1$ .

Given line is $x + 7 y + 2 = 0$ . Any line parallel to it is $x + 7 y + c = 0$ .

Distance from point $(1, - 1)$ to this line is $\frac{| 1 + 7 (- 1) + c |}{\sqrt{1^{2} + 7^{2}}} = 1 \Rightarrow | - 6 + c | = \sqrt{50}$

So $c - 6 = \pm \sqrt{50}$ and $c = 6 \pm \sqrt{50}$

Hence required lines are $x + 7 y + 6 + \sqrt{50} = 0$ and $x + 7 y + 6 - \sqrt{50} = 0$ .

Given line $3 x - 4 y - 5 = 0$ . Let required parallel lines be $3 x - 4 y + c = 0$ .

Distance between two parallel lines $\frac{| c + 5 |}{\sqrt{3^{2} + (- 4)^{2}}} = 1 \Rightarrow | c + 5 | = 5$ .

So $c + 5 = \pm 5$ and $c = 0$ or $c = - 10$

Hence required lines are $3 x - 4 y = 0$ and $3 x - 4 y - 10 = 0$ .

Let a line through $(0, a)$ be $y - a = m (x - 0)$ . So $y = m x + a$ .

Rewrite in standard form $m x - y + a = 0$ .

Distance from $(2 a, 2 a)$ is $\frac{| 2 a m - 2 a + a |}{\sqrt{m^{2} + 1}} = a$

$\Rightarrow (2 m - 1)^{2} = m^{2} + 1 \Rightarrow m = 0$ or $m = \frac{4}{3}$

Case 1: $m = 0$ gives $y = a$

Case 2: $m = \frac{4}{3}$ gives $y - a = \frac{4}{3} x \Rightarrow 4 x - 3 y + 3 a = 0$ .

From first $x = 3 y - 1$ . Substitute in second $2 (3 y - 1) + 5 y - 9 = 0 \Rightarrow 6 y - 2 + 5 y - 9 = 0 \Rightarrow y = 1 \Rightarrow x = 2$ .

So point of intersection is $(2, 1)$ . Let required line be $y - 1 = m (x - 2)$ . So $m x - y + (1 - 2 m) = 0$ .

Distance from origin is $\sqrt{5}$ . So $\frac{| 1 - 2 m |}{\sqrt{m^{2} + 1}} = \sqrt{5} \Rightarrow m = - 2$

Substitute into line equation $y - 1 = - 2 (x - 2) \Rightarrow y = - 2 x + 5$ .

From first line $x = y - 1$ . Substitute in second line $2 (y - 1) - 3 y + 5 = 0 \Rightarrow y = 3 \Rightarrow x = 2$

So the point is $(2, 3)$ . Let required line be $y - 3 = m (x - 2) \Rightarrow m x - y + (3 - 2 m) = 0$

Distance from $(3, 2)$ is $\frac{7}{5}$ , $\frac{| 3 m - 2 + 3 - 2 m |}{\sqrt{m^{2} + 1}} = \frac{7}{5} \Rightarrow \frac{| m + 1 |}{\sqrt{m^{2} + 1}} = \frac{7}{5}$

$\Rightarrow m = \frac{4}{3}$ or $m = \frac{3}{4}$

Case 1: $y - 3 = \frac{4}{3} (x - 2) \Rightarrow 4 x - 3 y + 1 = 0$ .

Case 2: $y - 3 = \frac{3}{4} (x - 2) \Rightarrow 3 x - 4 y + 6 = 0$ .

Given that the distance from $(1, 1)$ to the line $a x - b y + c = 0$ is $1$ .

So $\frac{| a - b + c |}{\sqrt{a^{2} + b^{2}}} = 1$

Squaring both sides $(a - b + c)^{2} = a^{2} + b^{2} \Rightarrow a^{2} + b^{2} + c^{2} - 2 a b + 2 a c - 2 b c = a^{2} + b^{2}$

$c^{2} - 2 a b + 2 a c - 2 b c = 0 \Rightarrow c^{2} + 2 a c - 2 b c = 2 a b$

Divide throughout by $2 a b c, \frac{c}{2 a b} + \frac{1}{b} - \frac{1}{a} = \frac{1}{c}$

Rearranging yields $\frac{1}{c} + \frac{1}{a} - \frac{1}{b} = \frac{c}{2 a}$ .

Given line is $\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$

So $b x \cos θ + a y \sin θ - a b = 0$

Distance from $(x_{1}, y_{1})$ is $\frac{| b x_{1} \cos θ + a y_{1} \sin θ - a b |}{\sqrt{b^{2} \cos^{2} θ + a^{2} \sin^{2} θ}}$

For $(\pm \sqrt{a^{2} - b^{2}}, 0)$

Product of perpendiculars $= \frac{| (b \sqrt{(a^{2} - b^{2})} \cos θ - a b) (- b \sqrt{a^{2} - b^{2}} \cos θ - a b) |}{b^{2} \cos^{2} θ + a^{2} \sin^{2} θ}$

$= \frac{| a^{2} b^{2} - b^{2} (a^{2} - b^{2}) \cos^{2} θ |}{b^{2} \cos^{2} θ + a^{2} \sin^{2} θ} = \frac{b^{2} (a^{2} \sin^{2} θ + b^{2} \cos^{2} θ)}{b^{2} \cos^{2} θ + a^{2} \sin^{2} θ} = b^{2}$ .

Given lines are $4 x + 3 y = 11$ and $8 x + 6 y = 15$ . Rewrite second line $8 x + 6 y = 15$ as $4 x + 3 y = \frac{15}{2}$ .

So these are two parallel lines. Distance between parallel lines $= \frac{| (- \frac{15}{2} - (- 11)) |}{\sqrt{4^{2} + 3^{2}}} = \frac{7}{10}$ .

Given lines are $2 x + 3 y = 19, 2 x + 3 y + 7 = 0$ and $2 x + 3 y = 6$ ; all have same slope.

Write in standard form $L_{1} : 2 x + 3 y - 19 = 0, L_{2} : 2 x + 3 y + 7 = 0$ and $L : 2 x + 3 y - 6 = 0$ .

Distance between parallel lines $a x + b y + c_{1} = 0$ and $a x + b y + c_{2} = 0$ is $\frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2}}}$

Distance of $L_{1}$ from $L = \frac{| - 19 - (- 6) |}{\sqrt{2^{2} + 3^{2}}} = \sqrt{13}$

Distance of $L_{2}$ from $L = \frac{| 7 - (- 6) |}{\sqrt{13}} = \sqrt{13}$

Hence, both distances are equal, so the lines are equidistant from $2 x + 3 y = 6$ .

Distance between parallel lines $= \frac{| c_{1} - c |}{\sqrt{m^{2} + 1}}$ .

Given sides are $3 x - 4 y = 0$ and $4 x + 3 y = 0$ .

These are perpendicular since $3.4 + (- 4) 3 = 0$ . So they are adjacent sides meeting at origin.

Area of square is $25$ , so side $= 5$ . Distance between each pair of parallel sides is $5$ .

For line $3 x - 4 y = 0$ , required parallel side is $3 x - 4 y + c = 0$

Distance from origin $\frac{| c |}{\sqrt{3^{2} + (- 4)^{2}}} = 5 \Rightarrow | c | = 25$

So lines are $3 x - 4 y + 25 = 0$ or $3 x - 4 y - 25 = 0$ .

For line $4 x + 3 y = 0$ , required parallel side is $4 x + 3 y + k = 0$ .

Distance from origin $\frac{| k |}{5} = 5 \Rightarrow | k | = 25$

So lines are $4 x + 3 y + 25 = 0$ or $4 x + 3 y - 25 = 0$ .

Taking consistent pair forming a square, the other two sides are

$3 x - 4 y + 25 = 0$ and $4 x + 3 y + 25 = 0$ .

Given lines are $L_{1} : a x + b y + c = 0, L_{2} : a_{1} x + b_{1} y + c = 0, L_{3} : a x + b y + c_{1} = 0$ , and $L_{4} : a_{1} x + b_{1} y + c_{1} = 0$

$L_{1}$ is parallel to $L_{3}$ and $L_{2}$ is parallel to $L_{4}$ , so they form a parallelogram.

Length of one pair of opposite sides equals distance between $L_{1}$ and $L_{3} = \frac{| c_{1} - c |}{\sqrt{a^{2} + b^{2}}}$

Length of the other pair equals distance between $L_{2}$ and $L_{4} = \frac{| c_{1} - c |}{\sqrt{a_{1}^{2} + b_{1}^{2}}}$

Given $a^{2} + b^{2} = a_{1}^{2} + b_{1}^{2}$ . So both lengths are equal.

Hence, all sides of parallelogram are equal. Therefore, it is a rhombus.

Given lines are $L_{1} : 4 x + 3 y - 6 = 0$ and $L_{2} : 5 x + 12 y + 9 = 0$ .

Angle bisectors are given by $\frac{4 x + 3 y - 6}{\sqrt{4^{2} + 3^{2}}} = \pm \frac{5 x + 12 y + 9}{\sqrt{5^{2} + 12^{2}}}$

$\Rightarrow \frac{4 x + 3 y - 6}{5} = \pm \frac{5 x + 12 y + 9}{13}$

Case I: $13 (4 x + 3 y - 6) = 5 (5 x + 12 y + 9) \Rightarrow 9 x - 7 y - 41 = 0$

Case II: $13 (4 x + 3 y - 6) = - 5 (5 x + 12 y + 9) \Rightarrow 7 x + 9 y - 3 = 0$

To identify angles, test origin $(0, 0)$

For $L_{1} : - 6 < 0$ . For $L_{2} : 9 > 0$ .

So origin lies between opposite signs, hence lies in one of the angles.

Checking bisectors: For $9 x - 7 y - 41 = 0 \Rightarrow - 41 < 0$

For $7 x + 9 y - 3 = 0 \Rightarrow - 3 < 0$

The bisector that preserves sign relation corresponds to angle containing origin.

So angle containing origin is $7 x + 9 y - 3 = 0$

Hence, acute angle bisector is $7 x + 9 y - 3 = 0$ and obtuse angle bisector is $9 x - 7 y - 41 = 0$ .

Given lines are $L_{1} : 3 x + 4 y - 5 = 0$ and $L_{2} : 12 x + 5 y - 7 = 0$ .

The locus of points equidistant from $L_{1}$ and $L_{2}$ is given by the angle bisectors:

$\frac{3 x + 4 y - 5}{5} = \pm \frac{12 x + 5 y - 7}{13}$ . So we get two bisectors.

Case I: $13 (3 x + 4 y - 5) = 5 (12 x + 5 y - 7) \Rightarrow 7 x - 9 y + 10 = 0$

Case II: $13 (3 x + 4 y - 5) = - 5 (12 x + 5 y - 7) \Rightarrow 99 x + 77 y - 100 = 0$

Now observe: The given line $7 x - 9 y + 10 = 0$ is exactly the angle bisector of $L_{1}$ and $L_{2}$ .

We have already proven this in Chapter 1, Section, Incenter of a triangle.

Given triangle sides $L_{1} : x + 1 = 0, L_{2} : 3 x - 4 y - 5 = 0$ , and $L_{3} : 5 x + 12 y - 27 = 0$ .

$A = L_{2} \cap L_{3}$ . From $3 x - 4 y = 5$ and $5 x + 12 y = 27$

$x = 3$ and $y = 1$ . So $A = (3, 1)$ .

$B = L_{3} \cap L_{1} \Rightarrow x = - 1$ and $y = \frac{8}{3}$ . So $B = (- 1, \frac{8}{3})$ .

$C = L_{1} \cap L_{2} \Rightarrow x = - 1$ and $y = - 2$ . So $C = (- 1, - 2)$ .

$B C = \sqrt{(0)^{2} + \frac{14^{2}}{3^{2}}} = \frac{14}{3}, C A = \sqrt{(4)^{2} + (3)^{2}} = 5$ , and $A B = \sqrt{(4)^{2} + (- 5 / 3)^{2})} = \frac{13}{3}$

Incenter formula is $I = \frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c}$ .

Hence incenter is $(\frac{1}{3}, \frac{2}{3})$ .

Let the given opposite sides be $x + y = 1$ and $x + y = 5$ .

Distance between the parallel sides is $d = \frac{| 5 - 1 |}{\sqrt{1^{2} + 1^{2}}} = 2 \sqrt{2}$

Let the side length of the rhombus be $a$ . Height of rhombus $= a \sin 45^{\circ}$

So $a \sin 45^{\circ} = 2 \sqrt{2} \Rightarrow a = 4$

Direction vector of lines $x + y = k$ is $(1, - 1)$ . Unit vector $= (\frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}})$

Vertex $A$ is $(2, - 1)$ . Adjacent vertex $B$ is $B = (2, - 1) \pm 4 * (\frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}}) \Rightarrow B = (2, - 1) \pm (2 \sqrt{2}, - 2 \sqrt{2})$

So $B_{1} = (2 + 2 \sqrt{2}, - 1 - 2 \sqrt{2})$ and $B_{2} = (2 - 2 \sqrt{2}, - 1 + 2 \sqrt{2})$

Second side makes $45^{\circ}$ with first side. Rotating $(1, - 1)$ by $45^{\circ}$ gives direction $(1, 0)$ .

So adjacent vertex $C$ is $C_{1} = (6, - 1)$ and $C_{2} = (- 2, - 1)$

Opposite vertex $D$ is given by $D = B + C - A$

Case I: $D = (2 + 2 \sqrt{2}, - 1 - 2 \sqrt{2}) + (6, - 1) - (2, - 1)$
$\Rightarrow D = (6 + 2 \sqrt{2}, - 1 - 2 \sqrt{2})$

Case II: $D = (2 - 2 \sqrt{2}, - 1 + 2 \sqrt{2}) + (- 2, - 1) - (2, - 1)$
$\Rightarrow D = (- 2 - 2 \sqrt{2}, - 1 + 2 \sqrt{2})$

Sides are parallel to $y = x + 2$ and $y = 7 x + 3$ . So slopes are $1$ and $7$ .

Let $A = (0, t)$ . Let adjacent vertices be $B = (x_{1}, y_{1})$ on line through $A$ with slope $1$ and $D = (x_{2}, y_{2})$ on line through $A$ with slope $7$

So $y_{1} - t = x_{1}$ and $y_{2} - t = 7 x_{2}$

Thus, $B = (x_{1}, t + x_{1})$ and $D = (x_{2}, t + 7 x_{2})$

For rhombus, adjacent sides are equal: $A B^{2} = A D^{2} \Rightarrow x_{1} = \pm 5 x_{2}$

Now $C = B + D - A = (x_{1} + x_{2}, t + x_{1} + 7 x_{2})$

Diagonals bisect each other at $(1, 2)$ i.e. Midpoint of $A C$ is $(1, 2)$

$((x_{1} + x_{2}) (2), * t + t + x_{1} + 7 x_{2}) (2)) = (1, 2)$

So $x_{1} + x_{2} = 2 \Rightarrow 2 t + x_{1} + 7 x_{2} = 4$

Case I: $x_{1} = 5 x_{2} \Rightarrow x_{2} = \frac{1}{3}$ and $x_{1} = \frac{5}{3}$

Then $2 t + 5 / 3 + \frac{7}{3} = 4 \Rightarrow t = 0$

So $A = (0, 0)$

Case II: $x_{1} = - 5 x_{2} \Rightarrow x_{2} = - \frac{1}{2}$ and $x_{1} = \frac{5}{2}$

Then $2 t + \frac{5}{2} - \frac{7}{2} = 4 \Rightarrow t = \frac{5}{2}$ .

So $A = (0, \frac{5}{2})$ .

We are given the lines $x - 2 y + 3 = 0$ and $4 x + 2 y - 5 = 0$ .

The angle bisectors are given by $\frac{a_{1} x + b_{1} y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2} |}} = \pm \frac{a_{2} x + b_{2} y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}$ .

For the given lines, $a_{1} = 1, b_{1} = - 2, c_{1} = 3$ and $a_{2} = 4, b_{2} = 2, c_{2} = - 5$ .

So, $\frac{x - 2 y + 3}{\sqrt{5}} = \pm \frac{4 x + 2 y - 5}{\sqrt{20}}$ .

Since $\sqrt{20} = 2 \sqrt{5}, \frac{x - 2 y + 3}{\sqrt{5}} = \pm \frac{4 x + 2 y - 5}{2 \sqrt{5}}$ .

Multiplying both sides by $\sqrt{5}, x - 2 y + 3 = \pm \frac{4 x + 2 y - 5}{2}$ .

Case I: $x - 2 y + 3 = \frac{4 x + 2 y - 5}{2} \Rightarrow 2 x + 6 y - 11 = 0$ .

Case II: $x - 2 y + 3 = - \frac{4 x + 2 y - 5}{2} \Rightarrow 6 x - 2 y + 1 = 0$ .

We want to prove that the line $6 x + 66 y - 7 = 0$ bisects the angle between the lines $15 x - 18 y - 1 = 0$ and $12 x + 10 y - 3 = 0$ .

The equation is $\frac{15 x - 18 y - 1}{\sqrt{15^{2} + (- 18)^{2}}} = \pm \frac{12 x + 10 y - 3}{\sqrt{12^{2} + 10^{2}}}$

$\sqrt{15^{2} + (- 18)^{2}} = \sqrt{549}$ and $\sqrt{12^{2} + 10^{2}} = \sqrt{244}$

So, $\frac{15 x - 18 y - 1}{\sqrt{549}} = \pm \frac{12 x + 10 y - 3}{\sqrt{244}}$

$\Rightarrow \sqrt{244} 15 x - 18 y - 1 = \pm \sqrt{549} (12 x + 10 y - 3)$

$\Rightarrow 30 x - 36 y - 2 = \pm (36 x + 30 y - 9)$

Taking plus sign we see that it is the given equation.

Rewrite the given lines in standard form: $24 x + 7 y - 20 = 0, 4 x - 3 y - 2 = 0$

Let $(x, y)$ be any point on $2 x + 11 y = 5$ .

Distance from first line: $D_{1} = \frac{| 24 x + 7 y - 20 |}{\sqrt{24^{2} + 7^{2}}} = \frac{| 24 x + 7 y - 20 |}{25}$

Distance from second line: $D_{2} = \frac{| 4 x - 3 y - 2 |}{\sqrt{4^{2} + (- 3)^{2}}} = \frac{| 4 x - 3 y - 2 |}{5}$

Now use the relation $2 x + 11 y = 5$ .

Then: $24 x + 7 y - 20 = (24 x + 132 y - 60) - 125 y + 40 = - 5 (25 y - 8)$

Also, $4 x - 3 y - 2 = (4 x + 22 y - 10) - 25 y + 8 = - (25 y - 8)$

Thus, $| 24 x + 7 y - 20 | = 5 | 25 y - 8 |$

$D_{1} = 5 \frac{| 25 y - 8 |}{25} = \frac{| 25 y - 8 |}{5}$ and $D_{2} = \frac{| 25 y - 8 |}{5}$ . Hence, $D_{1} = D_{2}$ .

We are given the lines $6 x + 8 y - 10 = 0$ and $4 x - 3 y - 7 = 0$ .

For a point $(x, y)$ equidistant from the two lines, $\frac{| 6 x + 8 y - 10 |}{\sqrt{6^{2} + 8^{2}}} = \frac{| 4 x - 3 y - 7 |}{\sqrt{4^{2} + (- 3)^{2}}}$

$\sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = 10$ and $\sqrt{4^{2} + (- 3)^{2}} = \sqrt{16 + 9} = 5$

So, $\frac{| 6 x + 8 y - 10 |}{10} = \frac{| 4 x - 3 y - 7 |}{5}$

Multiply: $| 6 x + 8 y - 10 | = 2 * | 4 x - 3 y - 7 | \Rightarrow 6 x + 8 y - 10 = \pm 2 (4 x - 3 y - 7)$

Case I: $6 x + 8 y - 10 = 8 x - 6 y - 14 \Rightarrow x - 7 y - 2 = 0$

Case II: $6 x + 8 y - 10 = - 8 x + 6 y + 14 \Rightarrow 7 x + y - 12 = 0$

Hence, the locus is $x - 7 y - 2 = 0$ or $7 x + y - 12 = 0$ .

The given lines are $x + y - 3 = 0$ and $7 x - y + 5 = 0$ .

For $x + y - 3 = 0$ we have $a_{1} = 1$ and $b_{1} = 1$ so the value is $\sqrt{2}$ .

For $7 x - y + 5 = 0$ we have $a_{2} = 7$ and $b_{2} = - 1$ so the value is $\sqrt{50} = 5 \sqrt{2}$ .

The angle bisectors satisfy $\frac{x + y - 3}{\sqrt{2}} = \pm \frac{7 x - y + 5}{5 \sqrt{2}}$

After simplification this becomes $5 (x + y - 3) = \pm (7 x - y + 5)$

First result $5 x + 5 y - 15 = 7 x - y + 5 \Rightarrow x - 3 y + 10 = 0$

Second result $5 x + 5 y - 15 = - 7 x + y - 5 \Rightarrow 6 x + 2 y - 5 = 0$

$\frac{3 x + 4 y - 11}{\sqrt{3^{2} + 4^{2}}} = \pm \frac{12 x - 5 y - 2}{\sqrt{12^{2} + (- 5)^{2}}}$ .

This simplifies to $\frac{3 x + 4 y - 11}{5} = \pm \frac{12 x - 5 y - 2}{13}$ .

Taking the negative sign gives $13 (3 x + 4 y - 11) = - 5 (12 x - 5 y - 2) \Rightarrow 11 x + 3 y - 17 = 0$ .

The slopes of the given lines are $- \frac{3}{4}$ and $\frac{12}{5}$ , which form an obtuse angle. Therefore, the bisector corresponding to the negative sign represents the acute angle.

Hence, the bisector of the acute angle is $11 x + 3 y = 17$ .

$\frac{x - 2 y + 4}{s q r t 5} = \pm \frac{4 x - 3 y + 2}{5}$ .

Taking the positive case gives $5 (x - 2 y + 4) = \sqrt{5} (4 x - 3 y + 2)$ .

Taking the negative case givesv $5 (x - 2 y + 4) = - \sqrt{5} (4 x - 3 y + 2)$ .

The slopes of the given lines are $\frac{1}{2}$ and $\frac{4}{3}$ , which form an acute angle. Therefore, the obtuse angle is the supplementary angle, and its bisector corresponds to the negative case.

Hence, the equation of the bisector of the obtuse angle is $5 (x - 2 y + 4) = - \sqrt{5} (4 x - 3 y + 2)$ .

The given lines are $x + y - 2 = 0$ and $x - y - 3 = 0$ .

These two lines divide the plane into four compartments depending on the signs of $x + y - 2$ and $x - y - 3$ .

Evaluating both expressions at each point.

For $(1, 0) : x + y - 2 = 1 + 0 - 2 = - 1 < 0$ and $x - y - 3 = 1 - 0 - 3 = - 2 < 0$ .

For $(2, 3) : x + y - 2 = 2 + 3 - 2 = 3 > 0$ and $x - y - 3 = 2 - 3 - 3 = - 4 < 0$ .

For $(1, - 4) : x + y - 2 = 1 - 4 - 2 = - 5 < 0$ and $x - y - 3 = 1 - (- 4) - 3 = 2 > 0$ .

For $(8, 1) : x + y - 2 = 8 + 1 - 2 = 7 > 0$ and $x - y - 3 = 8 - 1 - 3 = 4 > 0$ .

Each point gives a distinct sign combination for $(x + y - 2, x - y - 3)$ .

Thus, the four points lie in four different compartments formed by the given lines.

For $7 x - 5 y - 11 = 0 : 7 (0) - 5 (0) - 11 = - 11 < 0$ .

For $8 x + 3 y + 31 = 0 : 8 (0) + 3 (0) + 31 = 31 > 0$ .

For $x + 8 y - 19 = 0 : 0 + 8 (0) - 19 = - 19 < 0$ .

Thus, the origin lies on the negative side of the first line, the positive side of the second line, and the negative side of the third line.

Now consider a point clearly inside the triangle. Solving any two equations gives a vertex; testing a point between the vertices shows that the interior region corresponds to the same sign pattern $(-, +, -)$ .

Since the origin produces this same pattern, it lies inside the triangle.

Hence, the origin lies inside the triangle.

These lines are parallel and represent two opposite sides of the square.

The length of the side of the square is equal to the perpendicular distance between the two lines.

Thus, the distance between the lines is

$\frac{| - 65 - 26 |}{\sqrt{5^{2} + (- 12)^{2}}} = \frac{91}{\sqrt{25 + 144}} = 7$ .

So, the side of the square is $7$ . Hence, the area of the square is $7^{2} = 49$ .

The given sides of the square are $3 x + 4 y - 5 = 0$ and $3 x + 4 y - 15 = 0$ .

These two lines are parallel, so they represent opposite sides of the square.

So the side length is $\frac{| - 5 - (- 15) |}{\sqrt{3^{2} + 4^{2}}} = 2$ .

Since the third side passes through $(6, 5)$ and is perpendicular to the given sides, its equation has the form $4 x - 3 y + c = 0$ .

Substituting $(6, 5)$ gives $4 (6) - 3 (5) + c = 0 \Rightarrow c = - 9$ .

So one side of the square is $4 x - 3 y - 9 = 0$ .

The opposite side is parallel to it and is at a distance equal to the side length $2$ .

So the parallel line is $4 x - 3 y + k = 0$ .

Using distance formula between $4 x - 3 y - 9 = 0$ and $4 x - 3 y + k = 0$ ,

$\frac{| k + 9 |}{\sqrt{4^{2} + (- 3)^{2}}} = 2$ . Thus $k = 1$ or $k = - 19$ .

Since $(6, 5)$ lies on one side, the opposite side is $4 x - 3 y + 1 = 0$ .

Hence, the equations of the remaining two sides are $4 x - 3 y - 9 = 0$ and $4 x - 3 y + 1 = 0$ .

The given side of the rectangle is $3 x - 4 y - 10 = 0$ , and two vertices are $(2, 1)$ and $(2, 4)$ .

First, observe that the line joining $(2, 1)$ and $(2, 4)$ is $x = 2$ , since both points have the same $x$ -coordinate.

So one side of the rectangle lies on $x = 2$ .

Now find the distance between the parallel sides $x = 2$ and the given line $3 x - 4 y - 10 = 0$ .

Taking any point on $x = 2$ , for example $(2, 1)$ , the distance is

$\frac{| 3 (2) - 4 (1) - 10 |}{\sqrt{3^{2} + (- 4)^{2}}} = \frac{8}{5}$ .

So the height of the rectangle is $\frac{8}{5}$ .

The length of the rectangle is the distance between $(2, 1)$ and $(2, 4) : \sqrt{(2 - 2)^{2} + (4 - 1)^{2}} = 3$ .

Hence, the area of the rectangle is $3 \times \frac{8}{5} = \frac{24}{5}$ .

One diagonal joins $(2, 4)$ to the opposite vertex. The opposite side to $x = 2$ is parallel to it, so the opposite vertical line is $x = 2 + \frac{8}{5} = \frac{18}{5}$ or $x = 2 - \frac{8}{5} = \frac{2}{5}$ .

Since $(2, 1)$ lies on one side, the opposite vertex to $(2, 4)$ is $(\frac{18}{5}, 1)$ .

Now find the equation of the diagonal through $(2, 4)$ and $(\frac{18}{5}, 1)$ .

Slope is $\frac{1 - 4}{\frac{18}{5} - 2} = - \frac{15}{8}$ .

Equation is $y - 4 = - \frac{15}{8} (x - 2) \Rightarrow 15 x + 8 y - 62 = 0$ .

Hence, the area of the rectangle is $\frac{24}{5}$ and the required diagonal is $15 x + 8 y - 62 = 0$ .

The four lines are $a x + b y + c = 0, a x + b y - c = 0, a x - b y + c = 0$ , and $a x - b y - c = 0$ .

Opposite sides are parallel since each pair differs only in the constant term. Hence the figure formed is a parallelogram.

Intersect $a x + b y + c = 0$ and $a x - b y + c = 0$ .

We get $a x + b y = - c$ and $a x - b y = - c$ .

Adding gives $2 a x = - 2 c$ , so $x = - \frac{c}{a}$ . Substituting gives $y = 0$ . Thus one vertex is $(- \frac{c}{a}, 0)$ .

Similarly, other vertices are $(\frac{c}{a}, 0), (0, \frac{c}{b})$ , and $(0, - \frac{c}{b})$ .

The diagonals are the lines joining opposite vertices, so they lie along the coordinate axes.

Hence the diagonals are perpendicular, so the parallelogram is a rhombus.

Diagonal 1 has length $| \frac{c}{a} - (- \frac{c}{a}) | = \frac{2 | c |}{| a |}$ .

Diagonal 2 has length $| \frac{c}{b} - (- \frac{c}{b}) | = \frac{2 | c |}{| b |}$ .

Area of a rhombus is $\frac{1}{} 2 \times (p r o d u c t o f d i a g o n a l s)$ .

So area is $\frac{1}{2} \times \frac{2 | c |}{| a |} \times \frac{2 | c |}{| b |}$ .

This simplifies to $\frac{2 c^{2}}{| a b |}$ .

We choose a coordinate system such that the given straight line is the $x$ -axis. Then $p, q, r$ become the signed $y$ -coordinates of $A, B, C$ . So we write $A (x_{1}, p), B (x_{2}, q)$ , and $C (x_{3}, r)$ .

Figure 9.32.

Now we use the formula for the square of the area of a triangle in coordinate form: $4 Δ^{2} = (x_{1} (q - r) + x_{2} (r - p) + x_{3} (p - q))^{2} + (p (q - r) + q (r - p) + r (p - q))^{2}$ .

The second bracket simplifies as follows: $p (q - r) + q (r - p) + r (p - q) = 0$ .

So we get $4 Δ^{2} = (x_{1} (q - r) + x_{2} (r - p) + x_{3} (p - q))^{2}$ .

Now consider the squared side lengths using distance formula:

$a^{2} = (x_{2} - x_{3})^{2} + (q - r)^{2}, b^{2} = (x_{3} - x_{1})^{2} + (r - p)^{2}, c^{2} = (x_{1} - x_{2})^{2} + (p - q)^{2}$ .

Now we expand the required expression:

$a^{2} (p - q) (p - r) + b^{2} (q - r) (q - p) + c^{2} (r - p) (r - q)$ .

Substitute $a^{2}, b^{2}, c^{2} : [(x_{2} - x_{3})^{2} + (q - r)^{2}] (p - q) (p - r) + [(x_{3} - x_{1})^{2} + (r - p)^{2}] (q - r) (q - p) + [(x_{1} - x_{2})^{2} + (p - q)^{2}] (r - p) (r - q)$ .

We separate terms into two groups: those involving $x_{i}$ and those involving only $p, q, r$ .

The pure $p, q, r$ part simplifies to zero due to the identity $(p - q) (p - r) + (q - r) (q - p) + (r - p) (r - q) = 0$ .

So the expression reduces to $(x_{2} - x_{3})^{2} (p - q) (p - r) + (x_{3} - x_{1})^{2} (q - r) (q - p) + (x_{1} - x_{2})^{2} (r - p) (r - q)$ .

Now expand and regroup in terms of $x_{1}, x_{2}, x_{3}$ : This becomes $(x_{1} (q - r) + x_{2} (r - p) + x_{3} (p - q))^{2}$ .

But earlier we showed that $(x_{1} (q - r) + x_{2} (r - p) + x_{3} (p - q))^{2} = 4 Δ^{2}$ .

Hence, $a^{2} (p - q) (p - r) + b^{2} (q - r) (q - p) + c^{2} (r - p) (r - q) = 4 Δ^{2}$ .

Let the required line pass through $(4, - 5)$ with slope $m$ . Then its equation is $y + 5 = m (x - 4)$ , which gives $m x - y - 4 m - 5 = 0$ .

The distance of the point $(- 2, 3)$ from this line is $\frac{| m (- 2) - 3 - 4 m - 5 |}{\sqrt{m^{2} + 1}} = \frac{| 6 m + 8 |}{\sqrt{m^{2} + 1}}$ .

Given that this distance equals $12$ , we write $\frac{| 6 m + 8 |}{\sqrt{m^{2} + 1}} = 12$ .

Squaring both sides gives $(6 m + 8)^{2} = 144 (m^{2} + 1) \Rightarrow 27 m^{2} - 24 m + 20 = 0$ .

The discriminant is $D = (- 24)^{2} - 4 t i m e s 27 t i m e s 20 = 576 - 2160 = - 1584$ .

Since $D < 0$ , there is no real solution for $m$ . Hence, no line passing through $(4, - 5)$ can have distance $12$ from $(- 2, 3)$ .

Slope of $B C$ is $\frac{- 3 - (- 1)}{- 1 - (- 3)} = - 1$ .

So equation of $B C$ is $y + 1 = - 1 (x + 3) \Rightarrow x + y + 4 = 0$ .

A line parallel to $B C$ has the form $x + y + k = 0$ .

The perpendicular distance from origin $(0, 0)$ to this line is $\frac{| k |}{\sqrt{1^{2} + 1^{2}}} = \frac{| k |}{\sqrt{2}}$ .

Given this distance is $\frac{1}{2}$ , we write $\frac{| k |}{\sqrt{2}} = \frac{1}{2}$ .

So $| k | = \frac{\sqrt{2}}{2}$ , hence $k = \pm | f r a c s q r t 2 2$ .

Thus, the required lines are $x + y + \frac{\sqrt{2}}{2} = 0$ or $x + y - \frac{\sqrt{2}}{2} = 0$ .

Now check which line intersects segments $O B$ and $O C$ .

Line $O B$ : slope is $\frac{- 1 - 0}{- 3 - 0} = \frac{1}{3}$ , so equation is $y = \frac{1}{3} x$ .

Line $O C$ : slope is $\frac{- 3 - 0}{- 1 - 0} = 3$ , so equation is $y = 3 x$ .

On $O B$ , substitute $y = \frac{x}{3} : x + \frac{x}{3} + \frac{\sqrt{2}}{2} = 0$ gives a negative intersection point for $x$ , so it lies on segment $O B$ .

On $O C$ , substitute $y = 3 x : x + 3 x + \frac{\sqrt{2}}{2} = 0$ gives another valid intersection point on segment $O C$ .

Thus the required line is $x + y + \frac{\sqrt{2}}{2} = 0$ .

The center of the square is $C (1, - 1)$ and one side is $x - 2 y + 12 = 0$ .

Distance from $C$ to this line is $\frac{| 1 - 2 (- 1) + 12 |}{\sqrt{1^{2} + (- 2)^{2}}} = 3 \sqrt{5}$ .

So the side length of the square is $2 \times 3 \sqrt{5} = 6 \sqrt{5}$ .

The opposite side is parallel to the given line, so it is $x - 2 y + k = 0$ .

Distance between parallel sides equals $6 \sqrt{5}$ , so $\frac{| k - 12 |}{| s q r t 5} = 6 \sqrt{5}$ , giving $| k - 12 | = 30$ .

Thus $k = 42$ or $k = - 18$ , so opposite sides are $x - 2 y + 42 = 0$ and $x - 2 y - 18 = 0$ .

Now the other two sides are perpendicular to these, so have form $2 x + y + c = 0$ .

Distance from center gives $\frac{| 2 (1) - 1 + c |}{\sqrt{5}} = 3 \sqrt{5}$ , so $| 1 + c | = 15$ .

Thus, $c = 14$ or $c = - 16$ . Hence the remaining sides are $2 x + y + 14 = 0$ and $2 x + y - 16 = 0$ .

The given sides are $3 x - 2 y + 12 = 0$ and $x - 3 y + 11 = 0$ , and the diagonals intersect at $(2, 2)$ , which is the center of the parallelogram.

Opposite sides are parallel to the given ones.

For $3 x - 2 y + k = 0$ , using distance from $(2, 2) : \frac{| 3 (2) - 2 (2) + k |}{\sqrt{13}} = \frac{| 2 + k |}{\sqrt{13}}$ .

This equals the distance to $3 x - 2 y + 12 = 0$ , so $\frac{| 2 + k |}{=} 14$ , giving $k = 12$ or $k = - 16$ .

Hence other side is $3 x - 2 y - 16 = 0$ .

For $x - 3 y + k = 0 : \frac{| 2 - 6 + k |}{\sqrt{10}} = \frac{| k - 4 |}{\sqrt{10}}$ .

This equals the distance to $x - 3 y + 11 = 0$ , so $| k - 4 | = 7$ , giving $k = 11$ or $k = - 3$ .

Hence, other side is $x - 3 y - 3 = 0$ .

Intersect $3 x - 2 y + 12 = 0$ and $x - 3 y - 3 = 0$ to get $(- 6, - 3)$ . Diagonal through $(2, 2)$ and $(- 6, - 3)$ is $5 x - 8 y - 6 = 0$ .

Intersect $3 x - 2 y - 16 = 0$ and $x - 3 y + 11 = 0$ to get $(10, 7)$ . Diagonal through $(2, 2)$ and $(10, 7)$ is $5 x - 8 y + 6 = 0$ .

Hence, other sides are $3 x - 2 y - 16 = 0, x - 3 y - 3 = 0$ , and diagonals are $5 x - 8 y - 6 = 0, 5 x - 8 y + 6 = 0$ .

The given parallel lines are $3 x + 4 y + 2 = 0, 3 x + 4 y + 5 = 0$ , and $3 x + 4 y - 5 = 0$ .

Since all have the same $3 x + 4 y$ part, their relative positions depend only on constants $2, 5,$ and $- 5$ .

Clearly $- 5 < 2 < 5$ , so the line $3 x + 4 y + 2 = 0$ lies between $3 x + 4 y + 5 = 0$ and $3 x + 4 y - 5 = 0$ .

Now find the ratio in which it divides the distance between the other two lines.

So total distance between $3 x + 4 y + 5 = 0$ and $3 x + 4 y - 5 = 0$ is proportional to $| 5 - (- 5) | = 10$ .

Distance from $3 x + 4 y + 2 = 0$ to $3 x + 4 y + 5 = 0$ is proportional to $| 5 - 2 | = 3$ .

Distance from $3 x + 4 y + 2 = 0$ to $3 x + 4 y - 5 = 0$ is proportional to $| 2 - (- 5) | = 7$ .

Hence, the required ratio is $3 : 7$ .

Therefore, the line $3 x + 4 y + 2 = 0$ lies between the other two and divides their distance in the ratio $3 : 7$ .

The given lines are $x + 2 y + 3 = 0, x + 2 y - 7 = 0$ , and $2 x - y - 4 = 0$ .

The first two lines are parallel, so they form one pair of opposite sides of a square.

Their distance is the side length: $\frac{| 3 - (- 7) |}{\sqrt{1^{2} + 2^{2}}} = 2 \sqrt{5}$ .

Now the other pair of sides is parallel to $2 x - y - 4 = 0$ , so write $2 x - y + k = 0$ .

Distance between parallel sides is $2 \sqrt{5} : \frac{| k + 4 |}{\sqrt{2^{2} + (- 1)^{2}}} = 2 \sqrt{5}$ , so $\frac{| k + 4 |}{\sqrt{5}} = 2 \sqrt{5}$ , hence $| k + 4 | = 10$ .

Thus $k = 6$ or $k = - 14$ .

Therefore, the fourth sides are $2 x - y + 6 = 0$ or $2 x - y - 14 = 0$ .

The sides of the triangle are $3 x + 4 y - 6 = 0, 12 x - 5 y - 3 = 0$ , and $4 x - 3 y + 12 = 0$ .

Between $3 x + 4 y - 6 = 0$ and $12 x - 5 y - 3 = 0 : \frac{3 x + 4 y - 6}{5} = \frac{12 x - 5 y - 3}{13}$ , giving $13 (3 x + 4 y - 6) = 5 (12 x - 5 y - 3)$ , which simplifies to $9 x - 77 y + 63 = 0$ .

Between $12 x - 5 y - 3 = 0$ and $4 x - 3 y + 12 = 0 : \frac{12 x - 5 y - 3}{13} = \frac{4 x - 3 y + 12}{5}$ , giving $5 (12 x - 5 y - 3) = 13 (4 x - 3 y + 12)$ , which simplifies to $8 x + 19 y - 171 = 0$ .

Between $4 x - 3 y + 12 = 0$ and $3 x + 4 y - 6 = 0 : \frac{4 x - 3 y + 12}{5} = \frac{3 x + 4 y - 6}{5}$ , giving $4 x - 3 y + 12 = 3 x + 4 y - 6$ , which simplifies to $x - 7 y + 18 = 0$ .

Hence the internal bisectors of the triangle are $9 x - 77 y + 63 = 0, 8 x + 19 y - 171 = 0$ , and $x - 7 y + 18 = 0$ .

From $3 x + 4 y = 12$ and $5 x + 12 y = 20$ , we get $A = (4, 0)$ .

From $3 x + 4 y = 12$ and $- 7 x + 24 y = 22$ , we get $B = (2, 3 / 2)$ .

From $5 x + 12 y = 20$ and $- 7 x + 24 y = 22$ , we get $C = (\frac{18}{17}, \frac{125}{1 - 02})$ .

Now the side lengths are computed: $A B = \frac{5}{2}, B C = \frac{50}{51}, C A = 325 / 102$ .

The incenter is given by $I = \frac{a A + b B + c C}{a + b + c}$ where $a = B C, b = C A, c = A B$ .

So $a = \frac{50}{51}, b = \frac{325}{102}, c = \frac{5}{2}$ and $a + b + c = \frac{340}{51}$ .

$x = \frac{a .4 + b .2 + c . \frac{18}{17}}{\frac{340}{51}}$ . This simplifies to $x = \frac{33}{17}$ .

$y = \frac{a .0 + b . \frac{3}{2} + c . \frac{125}{102}}{\frac{340}{51}}$ . This simplifies to $y = \frac{20}{17}$ .

Let a point $(x, y)$ reflect to $(X, Y)$ in the line $x + y + 1 = 0$ .

For reflection in $a x + b y + c = 0$ , the formula gives

$X = x - 2 a \frac{a x + b y + c}{a^{2} + b^{2}}$ and $Y = y - 2 b \frac{a x + b y + c}{a^{2} + b^{2}}$ .

Here $a = 1, b = 1, c = 1$ , so $a^{2} + b^{2} = 2$ .

Thus, $X = x - (x + y + 1) = - y - 1$ and $Y = y - (x + y + 1) = - x - 1$

So the transformation is $x = - Y - 1, y = - X - 1$ .

Now the given line is $p x + q y + r = 0$ .

Substitute: $p (- Y - 1) + q (- X - 1) + r = 0 \Rightarrow q X + p Y + (p + q - r) = 0$

Hence, the reflection of the line is $q x + p y + (p + q - r) = 0$ .

The roads are $x - 2 y - 4 = 0$ and $2 x - y - 4 = 0$ .

Their intersection point is found from $x - 2 y = 4$ and $2 x - y = 4$ , giving $P = (4, 0)$ .

The direction of the angle bisector is obtained from normals $(1, - 2)$ and $(2, - 1)$ , giving $(3, - 3)$ , hence direction $(1, - 1)$ .

So the bisector through $P$ is $x + y - 4 = 0$ .

A direction vector is $(1, - 1)$ with magnitude $\sqrt{2}$ , so unit vector is $(\frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}})$ .

After moving $2$ km, the displacement is $(\sqrt{2}, - \sqrt{2})$ , so the point reached is $Q = (4 + \sqrt{2}, - \sqrt{2})$ .

The river bank is perpendicular to the path, so its direction is $(1, 1)$ .

Through $Q$ , its equation is $x - y - 4 - 2 \sqrt{2} = 0$ .

Thus, the river bank is $x - y - 4 - 2 \sqrt{2} = 0$ and the point of contact is $(4 + \sqrt{2}, - \sqrt{2})$ .

The sides of the rhombus are parallel to $y = 2 x + 3$ and $y = 7 x + 2$ , so their slopes are $2$ and $7$ .

Hence adjacent sides have slopes $2$ and $7$ , and the diagonals are along the angle bisectors of these directions.

The diagonals intersect at $(1, 2)$ , which is the midpoint of both diagonals.

So if a vertex is $A = (0, y)$ (since it lies on the $y$ -axis), its opposite vertex $C$ satisfies midpoint condition:

$\frac{0 + x_{C}}{2} = 1$ and $\frac{y + y_{C}}{2} = 2$ . So $x_{C} = 2$ and $y_{C} = 4 - y$ .

Thus, $C = (2, 4 - y)$ . Now slopes of sides must match $2$ and $7$ .

Take adjacent vertex $B$ from $A$ such that $A B$ has slope $2$ or $7$ .

Case I: slope of $A B = 2, \frac{y_{B} - y}{x_{B} - 0} = 2$

Since rhombus is symmetric about diagonals, solving consistently gives: $y = 1$

Case II: slope of $A B = 7$ . Similarly: $y = 3$

So possible vertices on $y$ -axis are: $(0, 1)$ and $(0, 3)$ .

Let two mutually perpendicular lines be $O X$ and $O Y$ . $A B$ be the variable line segment of constant length $l$ whose ends $A$ and $B$ move on the lines $O X$ and $O Y$ respectively.

Let $O A = a$ and $O B = b$ then $A = (a, 0)$ and $B = (0, b)$ .

Let $P$ divide the line segment in the ratio of $1 : 2$ . Let $P = (α, β)$ , then

$α = \frac{2 a}{3}$ and $β = \frac{b}{3}$ . Also, $l^{2} = O A^{2} + O B^{2} \Rightarrow a^{2} + b^{2} = l^{2}$

$9 α^{2} + 36 β^{2} = 4 l^{2}$ . Thus, locus of $P$ is $9 x^{2} + 36 y^{2} = 4 l^{2}$ .

Let the line cut the axes at $A = (p \sec α, 0)$ and $B = (0, p \csc α)$ . Let $P (h, k)$ be the middle point then $h = \frac{p}{2} \sec α, k = \frac{p}{2} \csc α$ .

Thus, $\cos α = \frac{p}{2 h}$ and $\sin α = \frac{p}{2 k}$

$\cos^{2} α + \sin^{2} α = \frac{p^{2}}{4 h^{2}} + \frac{p^{2}}{4 K^{2}} \Rightarrow \frac{1}{h^{2}} + \frac{1}{k^{2}} = \frac{4}{p^{2}}$

Hence, lcous of the point $P$ is $\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{4}{p^{2}}$ .

The point of intersection of the given lines is given by $(\frac{a b}{a + b}, \frac{a b}{a + b})$ .

Equation of line passing through this point is given by $y - \frac{a b}{a + b} = m (x - \frac{a b}{a + b})$ .

$\Rightarrow A = (\frac{a b (m - 1)}{m (a + b)}, 0)$ and $B = (0, \frac{a b (1 - m)}{a + b})$ .

Let $P (α, β)$ be the mid-point of $A B$ . We have to find its locus i.e. eliminate $m$ .

$α = - \frac{a b (1 - m)}{2 m (a + b)}$ and $β = \frac{a b (1 - m)}{2 (a + b)}$ .

$\frac{α}{β} = - \frac{1}{m} \Rightarrow m = - \frac{β}{α} \Rightarrow 2 α β (a + b) = a b (α + β) \Rightarrow 2 x y (a + b) = a b (x + y)$ .

Equation of any line perpendicular to the given equation passing through the origin is given by $\frac{x}{a} - \frac{y}{b} = 0$

Let foot of the perpendicular from the origin to the given line is intersection of the two lines. Let it be $P (α, β)$ , then

$\frac{α}{a} + \frac{β}{b} = 1$ and $\frac{α}{a} - \frac{β}{b} = 0$

Squaring and adding $α^{2} (\frac{1}{a^{2}} + \frac{1}{b^{2}}) + β^{2} (\frac{1}{b^{2}} + \frac{1}{a^{2}}) = 1 \Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = 1 / c^{2}$ .

Here $c$ is a constant and $a, b$ are parameters $\Rightarrow (α^{2} + β^{2}) . \frac{1}{c^{2}} = 1$ .

Hence, the locus of $P (α, β)$ is $x^{2} + y^{2} = c^{2}$ .

Equation of a straight line passing through $(h, k)$ is $y - k = m (x - h)$ .

Equation of the straight line perpendicular to the above line passing through origin is $y = - \frac{1}{m} x$ .

Let $P (α, β)$ be the foot of the perpendicular from the origin to first line. Clearly, $P (α, β)$ will be point of intersection of the two lines. Since $P$ lies on both the lines, therefore,

$β - k = m (α - x)$ and $β = - \frac{1}{m} α \Rightarrow m = - \frac{α}{β} =\Rightarrow y - k = - \frac{x}{y} (x - h) \Rightarrow x^{2} + y^{2} = h x + k y$ .

Let $O A = c$ , then $A = (0, c)$ , where $c$ is a constant. Let $C = (α, β)$ , then $O L = α = O B - B L = c \cot θ - B C \cos (θ + 60^{\circ})$

Figure 9.33.

$= c \cot θ - B C (\cos θ \cos 60^{\circ} - \sin θ \sin 60^{\circ}) = c \cot θ - \frac{A B}{2} \cos θ + \frac{A B}{2} \sqrt{3} \sin θ$

$= c \cot θ - \frac{1}{2} c \csc θ \cos θ + \frac{1}{2} c \csc θ \sin θ \sqrt{3}$

$= c \cot θ - \frac{c}{2} \cot θ + \frac{\sqrt{3} c}{2} = \frac{c}{2} (\cot θ + \sqrt{3})$

and $β = C L = C B \sin (θ + 60^{\circ}) = A B (\frac{1}{2} \sin θ + \cos θ \frac{\sqrt{3}}{2}) = \frac{c \csc θ}{2 (\sin θ + \sqrt{3} \cos θ)}$

$= \frac{c}{2} + \frac{\sqrt{3} c}{2} \cot θ$

Thus, we get $β = \sqrt{3} α - c$

Thus, locus of $C$ is $y = \sqrt{3} x - c$ , which is a straight line.

Let $P (x, y) . \Rightarrow \frac{| y - 2 x + 1 |}{\sqrt{5}} = \sqrt{x^{2} + y^{2}}$

$\Rightarrow (y - 2 x + 1)^{2} = 5 (x^{2} + y^{2}) \Rightarrow y^{2} + 4 x^{2} + 1 - 4 x y + 2 y - 4 x = 5 x^{2} + 5 y^{2}$

$\Rightarrow x^{2} + 4 y^{2} + 4 x y + 4 x - 2 y - 1 = 0$

For $y = 2 x : 25 x^{2} - 1 = 0 \Rightarrow x = \pm \frac{1}{5}$

Points: $Q (\frac{1}{5}, \frac{2}{5})$ and $R (- \frac{1}{5}, - 0 - \frac{2}{4})$

Midpoint: $= (0, 0)$ .

Equation of any line through origin can be written as $y = m x$ .

Solving it with the two equations we have $A = (\frac{2}{m + 2}, \frac{2 m}{m + 2})$ and $B = (\frac{2}{2 m - 1}, \frac{2 m}{2 m - 1})$ .

Let the coordinate of the mid-point be $(α, β)$ . Then $α = \frac{2 m + 1}{(2 m - 1) (m + 2)}$ and $β = \frac{m (3 m + 1)}{(2 m - 1) (m + 2)}$

$\Rightarrow \frac{α}{β} = \frac{1}{m} \Rightarrow m = \frac{β}{α}$

Thus, $α = \frac{(2 β + α) α}{(2 β - α) (β + 2 α)}$

Thus, locus is $2 x^{2} - 3 x y - 2 y^{2} + x + 3 y = 0$ .

Take $A (h, 0)$ and $B (- h, 0)$ . Let $O (x_{1}, y_{1})$ be a fixed point.

Let $P Q$ be a line through $P (a, b)$ and $Q (c, d)$ such that $O P ⊥ O Q$ .

So $(a - x_{1}) (c - x_{1}) + (b - y_{1}) (d - y_{1}) = 0$

Equation of line $P Q$ is: $(y - b) = m (x - a)$ where $m = \frac{d - b}{c - a}$ .

Foot of perpendicular from $O (x_{1}, y_{1})$ to $P Q$ is $R (x, y)$ , satisfying, $(x - x_{1}) + m (y - y_{1}) = 0$

Also $R$ lies on $P Q$ , so: $(y - b) = m (x - a)$

Eliminating $m$ gives: $(x - x_{1}) (c - a) + (y - y_{1}) (d - b) = 0$

Using $O P ⊥ O Q$ condition and simplifying yields: $(x - h) (x + h) + y^{2} = 0$

Hence: $x^{2} + y^{2} = h^{2}$ . This is a circle with center $(0, 0)$ and radius $h$ .

Therefore, the locus of $R$ is the circle with diameter $A B$ .

Figure 9.34.

Let $P Q$ be taken as $x$ -axis and its middle point is $O$ , the origin and $O Y$ as $y$ -axis.

Figure 9.35.

Let $P Q = 2 a$ then $P = (a, 0)$ and $Q = (- a, 0)$ . Let $R = (h, k)$ . Also let $∠ R P Q = θ$ and $∠ R Q P = ϕ$ .

Given that $θ - ϕ = 2 α$ . Let $R M$ be perpendicular to $x$ -axis.

Here $θ$ and $ϕ$ are variables and $a$ and $α$ are constants.

$\tan (θ - α) = \tan 2 α \Rightarrow \tan θ = \frac{k}{a - h}$ and $\tan ϕ = \frac{k}{a + h}$

Putting these values in $\tan (θ - α)$ yields

$h^{2} - k^{2} + 2 h k \cot 2 α - a^{2} = 0$

Hence, locus of $R$ is $x^{2} - y^{2} + 2 x y \cot 2 α - a^{2} = 0$ .

Equation of the line which passes through the point of intersection of the given lines is given by

$x + 2 y - 1 + k (2 x - y - 1) = 0$ , where $k$ is a parameter.

This line cuts $x$ and $y$ axes at $A = (\frac{k + 1}{2 k + 1}, 0)$ and $B = (0, \frac{k + 1}{2 - k})$ respectively.

Let $P (α, β)$ be the mid-point of $A B$ , then $α = (\frac{k + 1}{2 k + 1}, \frac{k + 1}{2 (2 - k)})$

$\frac{α}{β} = (2 - k) (2 k + 1) \Rightarrow k = \frac{2 β - α}{2 α + β}$

Putting $k$ back in $α$ we get $10 α β = 3 β + α \Rightarrow 10 x y = x + 3 y$ .

Let the variable line through $O$ make an angle $θ$ with the positive direction of $x$ -axis. Any point on this line will be $(r \cos θ + r \sin θ)$ .

Let the fixed lines be $y = m_{1} x + c_{1}$ and $y = m_{2} x + c_{2}$ .

Let $O R = r_{1}, O S = r_{2}$ and $O P = r_{3}$ , then according to question $\frac{m + n}{r_{3}} = \frac{m}{r_{1}} + \frac{n}{r_{2}}$

Since $A$ and $B$ lies on the fixed lines, therefore, $r_{1} \sin θ = m_{1} (r_{1} \cos θ) + c_{1}$ and $r_{2} \sin θ = m_{2} (r_{2} \cos θ) + c_{2}$

Let $P = (α, β)$ , then $α = r_{3} \cos θ, β = r_{3} \sin θ$

Thus, $\frac{m + n}{r_{3}} = \frac{m . (\sin θ - m_{1} \cos θ)}{c_{1}} + \frac{m_{2} . (\sin θ - m_{2} \cos θ)}{c_{2}}$

$\Rightarrow m + n = \frac{m}{c_{1}} (β - m_{1} α) + \frac{m}{c_{2}} (β - m_{2} α)$

The locus of $P$ is $m + n = (\frac{m}{c_{1}} + \frac{n}{c_{2}}) y - (\frac{m m_{1}}{c_{1}} + \frac{m m_{2}}{c_{2}}) x$ .

$\Rightarrow y - m_{1} x + c_{1} + \frac{n}{m} . \frac{c_{1}}{c_{2}} (y_{2} - m_{2} x - c_{2}) = 0$

Clearly locus of $P$ is a straight line passing through the point of intersection of the two fixed lines.

We take $O$ as the origin Let the $n$ lines be represented by $y = m_{r} x + c_{r}$ , where $r = 1, 2, 3, \dots, n$ .

Let $O A$ make an angle $θ$ with the $x$ -axis and cut the given $n$ lines at $n$ diferent points.

Let $O R_{1} = r_{1}, O R_{2} = r_{2}, \dots, O R_{n} = r_{n}, O R = r \Rightarrow R (α, β) = (r \cos θ, r \sin θ)$

$y = m_{n} x + c_{n} \Rightarrow r_{n} = \frac{c_{n}}{\sin θ - m_{n} \cos θ}$

$\Rightarrow \frac{n}{O R} = \frac{1}{O R_{1}} + \frac{1}{O R_{2}} + \dots + \frac{1}{O R_{n}}$

$\Rightarrow \frac{n}{r} = \frac{\sin θ - m_{1} \cos θ}{c_{1}} + \dots + \frac{\sin θ - m_{n} \cos θ}{c_{n}}$

$\Rightarrow n = \frac{β - m_{1} α}{c_{1}} + \dots + \frac{β - m_{n} α}{c_{n}} = (\frac{1}{c_{1}} + \frac{1}{c_{2}} + \dots + \frac{1}{c_{n}}) β - (\frac{m_{1}}{c_{1}} + \dots + \frac{m_{n}}{c_{m}}) α$

Hence, locus of $R$ is a straight line.

Let the base of the triangle $A B C$ be $B C$ which passes through a fixed point $P (f, g)$ . Let $B = (x_{1}, y_{1})$ and $C = (x_{2}, y_{2})$

Since $P, B, C$ are collinear, therefore, $| \begin{matrix} f & g & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{matrix} | = 0 \Rightarrow f (y_{1} - y_{2}) - g (x_{1} - x_{2}) + (x_{1} y_{2} - x_{2} y_{1}) = 0$

Let $A = (α, β)$ . We have to find the locus of the point $A$ .

Given lines are $y^{2} - 8 x y - 9 x^{2} = 0 \Rightarrow y - 9 x = 0$ and $x + y = 0$ .

Let these lines be perpendicular bisectors of $A B$ and $A C$ . Since $A B$ is perpendicular to line $y - 9 x = 0$ , therefore,

$\frac{β - y_{1}}{α - x_{1}} \times 9 = - 1 \Rightarrow x_{1} + 9 y_{1} = α + 9 β$

Since mid-point $(\frac{α + x_{1}}{2}, \frac{β + y_{1}}{2})$ of $A B$ lies on $y - 9 x = 0$ , therefore,

$(β + x_{1}) - 9 (β + y_{1}) = 0 \Rightarrow 9 x_{1} - y_{1} = β - 9 α$

Thus, $x_{1} = \frac{9 β - 40 α}{41}, y_{1} = \frac{40 β + 9 α}{41}$

Proceeding similarly for $A C$ we get $x_{2} = - β$ and $y_{2} = - α$

$\Rightarrow x_{1} - x_{2} = \frac{50 β - 40 α}{41}$ and $y_{1} - y_{2} = \frac{40 β + 50 α}{41}$ and $x_{1} y_{2} - x_{2} y_{1} = - \frac{1}{41} [α (9 β - 40 α) - β (40 β + 9 α)]$

Putting these in first equation we get locus of $A$ as

$4 (x^{2} + y^{2}) + (4 g + 5 f) x + (4 f - 5 g) y = 0$ .

Figure 9.36.

Let the coordinates of the vertex be $(h, k)$ , and let the lengths of the bases be respectively $l, l_{1}, l_{2}, \dots$ , and their equations be respectively

$x \cos α + y \sin α = p, x \cos β + y \sin β = p_{1}, \dots$ etc.

The length of the perpendiculars from $(h, k)$ on the bases will be

$| h \cos α + k \sin β - p |, | h \cos β + k \sin β - p_{1} |, \dots$ etc.

From the given condition sum of areas is constant. Thus,

$(x \sum l \cos α + y \sum l \sin α - \sum l p = k)$ , which is a straight line.

Let $A (\cos t, \sin t), B (\sin t, - \cos t), C (1, 2)$ .

Centroid $G (x, y)$ is given by $x = \frac{\cos t + \sin t + 1}{3}, y = \frac{\sin t - \cos t + 2}{3}$ .

So, $3 x - 1 = \cos t + \sin t, 3 y - 2 = \sin t - \cos t$ .

Now, $(\cos t + \sin t)^{2} + (\sin t - \cos t)^{2} = 2 (\sin^{2} t + \cos^{2} t) = 2$ .

Hence, $(3 x - 1)^{2} + (3 y - 2)^{2} = 2$ .

Given position: $x = u \cos α . t$ and $y = u \sin α . t - \frac{1}{2} g t^{2}$

From $x = u \cos α . t, t = \frac{x}{u \cos α}$

Substitute into $y : y = u \sin α . \frac{x}{u \cos α} - \frac{1}{2} g {(\frac{x}{u \cos α})}^{2}$

$\Rightarrow y = x \tan α - \frac{g x^{2}}{2 u^{2} \cos^{2} α}$

Hence, the locus is: $y = x \tan α - \frac{g x^{2}}{2 u^{2} \cos^{2} α}$

Let the line through $(1, 1)$ meet the axes at $A (a, 0)$ and $B (0, b)$ .

Equation of line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$

Since it passes through $(1, 1)$ , therefore, $\frac{1}{a} + \frac{1}{b} = 1$

Let midpoint of $A B$ be $M (x, y)$ . Then, $x = \frac{a}{2}, y = \frac{b}{2} \Rightarrow a = 2 x, b = 2 y$

Substitute into $\frac{1}{a} + \frac{1}{b} = 1$ to get $\frac{1}{2 x} + \frac{1}{2 y} = 1$

$\Rightarrow \frac{x + y}{2 x y} = 1 \Rightarrow x + y = 2 x y$

Hence, the locus is $2 x y = x + y$ .

Let the line through $(α, β)$ meet the axes at $A (a, 0)$ and $B (0, b)$ .

Equation in intercept form: $\frac{x}{a} + \frac{y}{b} = 1$ .

Since it passes through $(α, β)$ : $\frac{α}{a} + \frac{β}{b} = 1$ .

Let midpoint be $M (x, y)$ with $x = \frac{a}{2}, y = \frac{b}{2}$ . So $a = 2 x, b = 2 y$ .

Substitute to get $\frac{α}{2 x} + \frac{β}{2 y} = 1$ .

Hence, $\frac{α}{x} + \frac{β}{y} = 2$ . Multiplying by $x y$ gives $α y + β x = 2 x y$ .

Let the line cut the axes at $A (a, 0)$ and $B (0, b)$ .

Then its intercept form is $\frac{x}{a} + \frac{y}{b} = 1$

Given condition is $a + b = k$

Midpoint of $A B$ is $M (x, y)$ , so $x = \frac{a}{2}, y = \frac{b}{2} \Rightarrow a = 2 x, b = 2 y$

Substitute into $a + b = k$ gives $2 x + 2 y = k$

Hence, the locus is $x + y = \frac{k}{2}$ .

Let the line meet the axes at $A (x_{1}, 0)$ and $B (0, y_{1})$ .

Since $A P = b$ and $P B = a$ , the total length is constant, therefore $A B = a + b$

Point $P (x, y)$ divides $A B$ internally in the ratio, thus, $A P : P B = b : a$

Using section formula: $x = \frac{b . o + a . x_{1}}{a + b} = \frac{a x_{1}}{a + b}, y = \frac{b . y_{1} - a .0}{a + b}$

So, $x_{1} = \frac{(a + b) x}{a}$ and $y_{1} = \frac{(a + b) y}{b}$

Since $A$ and $B$ are intercepts of the same line, therefore, $\frac{x_{1}}{a} + \frac{y_{1}}{b} = 1$

Substituting values yields $\frac{(a + b) x}{a^{2}} + \frac{(a + b) y}{b^{2}} = 1$

Dividing by $(a + b)$ gives $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ .

Let the variable line cut the axes at $A (a, 0)$ and $B (0, b)$ .

Since the line passes through $(\frac{6}{5}, \frac{6}{5})$ , therefore $\frac{x}{a} + \frac{y}{b} = 1$

Substituting $(\frac{6}{5}, \frac{6}{5})$ yields $\frac{6}{5 a} + \frac{6}{5 b} = 1 \Rightarrow 6 (a + b) = 5 a b$

Point $P (x, y)$ divides $A B$ internally in ratio $2 : 1 :\Rightarrow A P : P B = 2 : 1$

Using section formula: $x = \frac{2.0 + 1. a}{2 + 1} = \frac{a}{3}$ and $y = \frac{2. b + 1.0}{2 + 1} = \frac{2 b}{3}$

So, $a = 3 x, b = \frac{3 y}{2}$

Substitute into $6 (a + b) = 5 a b$ gives us $6 (3 x + \frac{3 y}{2}) = 5.3 x . \frac{3 y}{2}$

Hence, $5 x y = 2 (2 x + y)$ .

Let the moving line cut the axes at $A (a, 0)$ and $B (0, b)$ .

Equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$

Perpendicular distance from origin is given as $p$ is $\frac{| a b |}{\sqrt{a^{2} + b^{2}}} = p$

Squaring yields $a^{2} b^{2} = p^{2} (a^{2} + b^{2})$

Centroid of triangle $O A B$ is $G (x, y) : x = \frac{a}{3}, y = \frac{b}{3} \Rightarrow a = 3 x, b = 3 y$

Substitute into distance condition to get $(9 x^{2}) (9 y^{2}) = p^{2} (9 x^{2} + 9 y^{2})$

$81 x^{2} y^{2} = 9 p^{2} (x^{2} + y^{2})$ . Hence the locus of centroid is $9 x^{2} y^{2} = p^{2} (x^{2} + y^{2})$ .

Let $P (x, y)$ be the moving point. Given $A P ⊥ B P$ .

Slope of $A P = \frac{y - y_{1}}{x - x_{1}}$ and slope of $B P = \frac{y - y_{2}}{x - x_{2}}$

Since lines are perpendicular, therefore, $\frac{y - y_{1}}{x - x_{1}} * \frac{y - y_{2}}{x - x_{2}} = - 1$

$\Rightarrow (y - y_{1}) (y - y_{2}) = - (x - x_{1}) (x - x_{2}) \Rightarrow (x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$

Hence, the locus is $(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$ .

Let $P (x, y)$ be the moving point.

Square of distance from $(3, - 2)$ is $(x - 3)^{2} + (y + 2)^{2}$

Distance from line $5 x - 12 y = 13$ is $\frac{| 5 x - 12 y - 13 |}{\sqrt{5^{2} + (- 12)^{2}}} = \frac{| 5 x - 12 y - 13 |}{13}$

Given condition is $(x - 3)^{2} + (y + 2)^{2} = \frac{| 5 x - 12 y - 13 |}{13}$

Multiply both sides by $13$ gives $13 [(x - 3)^{2} + (y + 2)^{2}] = | 5 x - 12 y - 13 |$

Squaring both sides yields $169 {[(x - 3)^{2} + (y + 2)^{2}]}^{2} = (5 x - 12 y - 13)^{2}$ .

Let $P (x, y)$ be the moving point. Fixed points are $F_{1} (a e, 0)$ and $F_{2} (- a e, 0)$ .

Given $\sqrt{(x - a e)^{2} + y^{2}} + \sqrt{(x + a e)^{2} + y^{2}} = 2 a$

This is the definition of an ellipse with foci $F_{1}, F_{2}$ and constant sum $2 a$ .

Squaring and simplifying gives $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} (1 - e^{2})} = 1$ .

Let the line $y = x + c$ intersect $2 x + 3 y = 5$ at $A (x_{1}, y_{1})$ and $2 x + 3 y = 8$ at $B (x_{2}, y_{2})$ .

For $A : y = x + c$ . Substituting in $2 x + 3 y = 5$ gives $2 x + 3 (x + c) = 5 \Rightarrow 5 x + 3 c = 5$

$\Rightarrow x_{1} = \frac{5 - 3 c}{5}, y_{1} = x_{1} + c = \frac{5 + 2 c}{5}$

For $B : 2 x + 3 (x + c) = 8 \Rightarrow 5 x + 3 c = 8 \Rightarrow x_{2} = \frac{8 - 3 c}{5}, y_{2} = \frac{8 + 2 c}{5}$

Midpoint $M (x, y) : x = \frac{x_{1} + x_{2}}{2} = \frac{13 - 6 c}{10}$ and $y = \frac{y_{1} + y_{2}}{2} = \frac{13 + 4 c}{10}$

Eliminating $c$ From $x = \frac{13 - 6 c}{10} \Rightarrow 6 c = 13 - 10 x$

Substitute into $y$ yields $y = (\frac{13 + 4 \frac{13 - 10 x}{6}}{10}) \Rightarrow 4 x + 6 y = 13$ .

Let the point $A$ be $(a, 0)$ on the $x$ -axis and the point $B$ be $(b, 6 b)$ on the line $y = 6 x$ .

The length of $A B$ is $2 l$ , so using the distance formula we get $(b - a)^{2} + (6 b)^{2} = 4 l^{2}$ .

The midpoint of $A B$ is $M (x, y)$ , so $x = \frac{a + b}{2}$ and $y = \frac{0 + 6 b}{2}$ .

From this we obtain $b = \frac{y}{3}$ and $a = 2 x - \frac{y}{3}$ .

Substituting these values into the length condition gives ${(\frac{y}{3} - (2 x - \frac{y}{3}))}^{2} + 36 {(\frac{y}{3})}^{2} = 4 l^{2}$ .

This simplifies to ${(\frac{2 y}{3} - 2 x)}^{2} = 4 l^{2}$ .

Dividing by $4$ gives ${(\frac{y}{3} - x)}^{2} + y^{2} = l^{2}$ .

Let the variable line through $P (- 1, 2)$ meet the $x$ -axis at $A (a, 0)$ and the $y$ -axis at $B (0, b)$ .

The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$ .

Since it passes through $P (- 1, 2)$ we get $- \frac{1}{a} + \frac{2}{b} = 1$ .

This gives the relation $- b + 2 a = a b$ .

Let $Q (x, y)$ lie on $A B$ . Then $Q$ divides $A B$ in some ratio $t : 1$ .

Using section formula we get $x = \frac{a}{1 + t}$ and $y = \frac{t b}{1 + t}$ .

Hence $a = x (1 + t)$ and $b = \frac{y (1 + t)}{t}$ .

Since $P, A, Q, B$ are such that $P A, P Q, P B$ are in harmonic progression we use $P Q^{- 1} = \frac{P A^{- 1} + P B^{- 1}}{2}$ .

This gives the relation $\frac{1}{P Q} = \frac{\frac{1}{P A} + \frac{1}{P B}}{2}$ .

Using coordinate distances we obtain after simplification that $y = 2 x$ .

Hence the locus of $Q$ is the line $y = 2 x$ .

$\frac{1}{P A} + \frac{1}{P B} = \frac{2}{P Q} \Rightarrow y = 2 x$

$- \frac{1}{P A} - \frac{1}{P B} = \frac{2}{P Q} \Rightarrow y = 2 x + 8$

$\frac{1}{P A} - \frac{1}{P B} = \frac{2}{P Q} \Rightarrow y = - 2 x + 4$

$- \frac{1}{P A} + \frac{1}{P B} = \frac{2}{P Q} \Rightarrow y = - 2 x - 4$

Hence, the locus is a rhombus bounded by these four lines excluding the vertices.

Let $B (x_{1}, y_{1})$ be a variable point.

The perpendicular bisector of $O B$ passes through the midpoint of $O (0, 0)$ and $B (x_{1}, y_{1})$ , which is $(\frac{x_{1}}{2}, \frac{y_{1}}{2})$ , and has slope perpendicular to $O B$ .

Equation of perpendicular bisector of $O B$ is $x_{1} x + y_{1} y = \frac{x_{1}^{2} + y_{1}^{2}}{2}$

The midpoint of $A B$ , where $A (4, 4)$ and $B (x_{1}, y_{1})$ , is $(\frac{x_{1} + 4}{2}, \frac{y_{1} + 4}{2})$ .

Slope of $A B$ is $\frac{y_{1} - 4}{x_{1} - 4}$ , so the perpendicular bisector of $A B$ is $(x_{1} - 4) (x - \frac{x_{1} + 4}{2}) + (y_{1} - 4) (y - \frac{y_{1} + 4}{2}) = 0$

The point of intersection $P (x, y)$ satisfies both equations.

Eliminating $x_{1}$ and $y_{1}$ from the two equations gives a relation between $x$ and $y$ .

After simplification, we obtain $(x - 2)^{2} + (y - 2)^{2} = 8$ .

Let $O$ be the origin and let $A (a, 0)$ and $B (0, b)$ be fixed points on the axes.

Let $C (c, 0)$ and $D (0, d)$ be variable points on the axes.

The given condition is $\frac{1}{c} - \frac{1}{d} = \frac{1}{a} - \frac{1}{b}$ .

Let $P (x, y)$ be the point of intersection of the lines $A D$ and $B C$ .

The equation of the line $A D$ is $y = d (1 - \frac{x}{a})$ .

The equation of the line $B C$ is $y = b - \frac{b}{c} x$ .

At the intersection point $P (x, y)$ we have $d (1 - \frac{x}{a}) = b - \frac{b}{c} x$ .

This simplifies to $d - \frac{d}{a} x = b - \frac{b}{c} x$ .

Rearranging gives $(b - d) = x (\frac{b}{c} - \frac{d}{a})$ .

Using the given condition $\frac{1}{c} - \frac{1}{d} = \frac{1}{a} - \frac{1}{b}$ we rewrite it as $\frac{d - c}{c d}) = \frac{b - a}{a b}$ .

This gives $a b (d - c) = c d (b - a)$ .

On solving the system and eliminating $c$ and $d$ we obtain $x = y$ .

Therefore, the locus of the point $P$ is $x - y = 0$ .

Let $Q = (a, k)$ . Since $Q R$ is the bisector of $∠ O Q A$ , therefore,

$\frac{O R}{R A} = \frac{O Q}{Q A} = \frac{\sqrt{a^{2} + k^{2}}}{k} ∴ R = (\frac{a \sqrt{a^{2} + k^{2}}}{\sqrt{a^{2} + k^{2}} + k}, 0)$

Let $P (α, β)$ be the foot of the perpendicular from $R$ to $O Q$ .

Now $Q, P, Q$ are collinear. $∴ | \begin{matrix} 0 & 0 & 1 \\ α & β & 1 \\ a & k & 1 \end{matrix} | = 0 \Rightarrow k = \frac{a β}{α}$

Again $R P ⊥ Q Q \Rightarrow \frac{β}{a - \frac{a \sqrt{a^{2} + k^{2}}}{\sqrt{a^{2} + k^{2}} + k}} . \frac{k}{a} = - 1$

Putting the value of $k$ we get locus as ${(a x - x^{2} - y^{2})}^{2} (a^{2} y^{2} + x^{2}) = a^{2} y^{2} {(x^{2} + y^{2})}^{2}$ .

As shown in the diagram $(h - a)^{2} + k^{2} = a^{2} = C B^{2}, h^{2} + (k - b)^{2} = a^{2} = C A^{2}$

We have two equations and two unknowns. Solving thes gives locus of $C (h, k)$ as

$b x \pm a y = 0$ .

Figure 9.37.

Let the isosceles triangle have base $B C$ on the $x$ -axis with $B (- a, 0)$ and $C (a, 0)$ , and vertex $A (0, h)$ .

Let $P (x, y)$ be the moving point.

The distance from $P$ to the base $B C$ is $y$ , so the square of the distance is $y^{2}$ .

The equation of line $A B$ is $h x + a y - a h = 0$ , so the distance from $P$ to $A B$ is $\frac{| h x + a y - a h |}{\sqrt{h^{2} + a^{2}}}$ .

The equation of line $A C$ is $h x - a y + a h = 0$ , so the distance from $P$ to $A C$ is $\frac{| h x - a y + a h |}{\sqrt{h^{2} + a^{2}}}$ .

The given condition is $y^{2} = \frac{| h x + a y - a h |}{\sqrt{h^{2} + a^{2}}} \times \frac{| h x - a y + a h |}{\sqrt{h^{2} + a^{2}}}$ .

This simplifies to $y^{2} (h^{2} + a^{2}) = (h x + a y - a h) (h x - a y + a h)$ .

Expanding the product gives $h^{2} x^{2} - a^{2} y^{2} + a^{2} h^{2}$ .

Substituting gives $y^{2} (h^{2} + a^{2}) = h^{2} x^{2} - a^{2} y^{2} + a^{2} h^{2}$ .

Rearranging gives $h^{2} x^{2} + a^{2} h^{2} = y^{2} (h^{2} + 2 a^{2})$ .

Dividing by $h^{2}$ gives $x^{2} + a^{2} - y^{2} - \frac{2 a^{2}}{h^{2}} y^{2} = 0$ .

Rearranging into standard form gives $x^{2} + y^{2} (1 + \frac{2 a^{2}}{h^{2}}) = a^{2}$ .

Therefore, the locus of $P$ is a circle.

Let $O$ be the origin. A variable line through $O$ meets two fixed straight lines at $R$ and $S$ .

Let $P$ lie on this line such that $\frac{2}{O P} = \frac{1}{O R} + \frac{1}{O S}$ .

Since $O, R, S, P$ are collinear, let $O R = r, O S = s$ , and $O P = p$ .

The given condition is $\frac{2}{p} = \frac{1}{r} + \frac{1}{s}$ .

This gives $2 = p (\frac{1}{r} + \frac{1}{s})$ .

Multiplying by $r s$ gives $2 r s = p (r + s)$ .

Hence $p = \frac{2 r s}{r + s}$ .

Let the variable line through $O$ have direction ratio $(1, m)$ so that $R = (r, m r), S = (s, m s)$ , and $P = (p, m p)$ .

The points $R$ and $S$ lie on two fixed straight lines, so $r$ and $s$ satisfy linear equations depending on $m$ .

Eliminating $r$ and $s$ between these two linear relations and the condition $2 r s = p (r + s)$ removes the parameter $m$ .

The resulting relation between $x$ and $y$ is linear.

Therefore, the locus of $P$ is a straight line.

The $i$ -th terms are $x_{i} = p + (i - 1) a$ and $y_{i} = q + (i - 1) b$ .

The mean of the first $n$ terms of an arithmetic sequence is the average of first and last terms.

Hence $α = \frac{x_{1} + x_{n}}{2}$ .

Substituting gives $α = \frac{p + p + (n - 1) a}{2}$ .

So $α = p + \frac{(n - 1) a}{2}$ .

Similarly $β = \frac{y_{1} + y_{n}}{2} = q + \frac{(n - 1) b}{2}$ .

Eliminating $n$ from both equations gives $\frac{α - p}{a} = \frac{β - q}{b}$ .

Hence, $b (α - p) = a (β - q)$ i.e. the locus is $b x - a y + q a - b p = 0$ .

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Chapter 9. Answers of Straight Lines