Chapter 7. Answers of Coordinates

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The area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ 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 given points, we compute
$Δ = \frac{1}{2} | 1 (6 - (- 1)) + (- 7) (- 1 - 3) + 5 (3 - 6) |$

This simplifies to $Δ = \frac{1}{2} | 7 + 28 - 15 | = 10$ .

Substituting the given points in the formula, we obtain
$Δ = \frac{1}{2} | 0 (6 - (- 2)) + 3 ((- 2) - 4) + (- 8) (4 - 6) |$

This simplifies to $Δ = \frac{1}{2} | - 2 | = 1$ .

ubstituting the given points in the formula, we obtain
$Δ = \frac{1}{2} | 5 ((- 3) - (- 5)) + (- 9) ((- 5) - 2) + (2 - (- 3)) |$

Simplifying each term, $Δ = \frac{1}{2} | 10 + 63 - 15 | = 29$ .

$Δ = \frac{1}{2} | a ((b - c) - c) + a (c - (b + c)) + (- a) ((b + c) - (b - c)) |$

Simplifying each term, $Δ = \frac{1}{2} | a (b - 2 c) - a b - 2 a c | = \frac{1}{2} | - 4 a c | = 2 a c$ .

$Δ = \frac{1}{2} | a \sin ϕ_{1} (a \sin ϕ_{2} - a \sin ϕ_{3}) + a \cos ϕ_{2} (a \sin ϕ_{3} - a \sin ϕ 1) + a \cos ϕ_{3} (a \sin ϕ_{1} - a \sin ϕ_{2}) |$

$= \frac{a^{2}}{2} | \sin (ϕ_{2} - ϕ_{1}) + \sin (ϕ_{3} - ϕ_{2}) + \sin (ϕ_{1} - ϕ_{3}) |$ .

$Δ = \frac{1}{2} | a m_{1}^{2} (2 a m_{2} - 2 a m_{3}) + a m_{2}^{2} (2 a m_{3} - 2 a m_{1}) + a m_{3}^{2} (2 a m_{1} - 2 a m_{2}) |$

$Δ = a^{2} | m_{1}^{2} m_{2} - m_{1}^{2} m_{3} + m_{2}^{2} m_{3} - m_{2}^{2} m_{1} + m_{3}^{2} m_{1} - m_{3}^{2} m_{2} |$

$Δ = a^{2} | (m_{1} - m_{2}) (m_{2} - m_{3}) (m_{3} - m_{1}) |$ .

$Δ = \frac{1}{2} | 1 ((- 2) - 16) + 3 (16 - 4) + (- 3) (4 - (- 2)) |$

$Δ = \frac{1}{2} | - 18 + 36 - 18 | = 0$ .

Since the area is zero, the points $A, B$ , and $C$ lie on a straight line and are collinear.

$Δ = \frac{1}{2} | (- \frac{1}{2}) (6 - 8) + (- 5) (8 - 3) + (- 8) (3 - 6) | = 0$

Since the area of the triangle is zero, therefore, the points are collinear.

$Δ = \frac{1}{2} | a ((c + a) - (a + b)) + b ((a + b) - (b + c)) + c ((b + c) - (c + a)) | = 0$

Hence, the area is zero, which proves that the points collinear.

Let $A B C$ be a triangle with vertices $A, B$ and $C$ . Let $D$ be the midpoint of $B C$ . Then the coordinates of $D$ are $(\frac{x B + x C}{2}, \frac{y B + y C}{2})$ .

$A B^{2} = (x B - x A)^{2} + (y B - y A)^{2}$ , and $A C^{2} = (x C - x A)^{2} + (y C - y A)^{2}$

$C D^{2} = {(x C - \frac{x B + x C}{2})}^{2} + {(y C - \frac{y B + y C}{2})}^{2} = {(\frac{x C = x C}{2})}^{2} + {(\frac{y C - y B}{2})}^{2} = \frac{1}{4} B C^{2}$

Now, consider the sum $A B^{2} + A C^{2}$ . Using the midpoint formula, we can write $A C^{2} + A D^{2} + C D^{2} = A B^{2} + \frac{A C^{2}}{2}$ , and after simplification, we obtain $A B^{2} + A C^{2} = 2 (A D^{2} + C D^{2})$

Hence, in any triangle $A B C$ , if $D$ is the midpoint of $B C$ , then $A B^{2} + A C^{2} = 2 (A D^{2} + C D^{2})$ .

Let triangle $A B C$ have vertices $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ .

Let $D$ , $E$ and $F$ be the midpoints of $B C, C A$ and $A B$ respectively. Then $D = (\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2}), E = (\frac{x_{3} + x_{1}}{2}, \frac{y_{3} + y_{1}}{2}), F = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

Let $P$ be the point that divides $A D$ internally in the ratio $2 : 1$ . By the section formula, $P = (\frac{1. x_{1} + 2. \frac{x_{2} + x_{3}}{2}}{1 + 2}, \frac{1. y_{1} + 2. \frac{y_{2} + y_{3}}{2}}{1 + 2})$ .

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

Similarly, for the line $B E$ , let $Q$ divide it internally in the ratio $2 : 1$ . Then $Q = (\frac{1. x_{2} + 2. \frac{x_{3} + x_{1}}{2}}{1 + 2}, \frac{1. y_{2} + 2. \frac{y_{3} + y_{1}}{2}}{1 + 2})$ ,

which simplifies to $Q = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ .

For the line $C F$ , let $R$ divide it internally in the ratio $2 : 1$ . Then $R = (\frac{1. x_{3} + 2. \frac{x_{1} + x_{2}}{2}}{1 + 2}, \frac{1. y_{3} + 2. \frac{y_{1} + y_{2}}{2}}{1 + 2})$ ,

which simplifies to $R = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ .

Hence, the point that divides $A D$ in the ratio $2 : 1$ also divides $B E$ and $C F$ in the same ratio. This point is the centroid of triangle $A B C$ .

Let the vertices of the quadrilateral be $A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})$ and $D (x_{4}, y_{4})$ .

The area of a quadrilateral can be computed by dividing it into two triangles, for example $A B C$ and $A C D$ , and adding their areas.

The area of triangle $A B C$ is $\frac{1}{2} | x_{1} * (y_{2} - y_{3}) + x_{2} * (y_{3} - y_{1}) + x_{3} * (y_{1} - y_{2}) |$ .

The area of triangle $A C D$ is $\frac{1}{2} | x_{1} * (y_{3} - y_{4}) + x_{3} * (y_{4} - y_{1}) + x_{4} * (y_{1} - y_{3}) |$ .

Hence, the area of the quadrilateral is
$◻ A B C D = \frac{1}{2} | x_{1} * (y_{2} - y_{3}) + x_{2} * (y_{3} - y_{1}) + x_{3} * (y_{1} - y_{2}) | + \frac{1}{2} | x_{1} * (y_{3} - y_{4}) + x_{3} * (y_{4} - y_{1}) + x_{4} * (y_{1} - y_{3}) |$ .

Substituting the given coordinates in the formula obtained in previous problem, we get

$Δ = \frac{1}{2} * | 1 * 4 + 3 * (- 2) + 5 * (- 7) + 4 * 1 - (1 * 3 + 4 * 5 + (- 2) * 4 + (- 7) * 1) |$

Simplifying each term, we get $Δ = \frac{1}{2} * | 4 - 6 - 35 + 4 - (3 + 20 - 8 - 7) |$

$Δ = \frac{1}{2} * | - 33 - (- 12) | = \frac{1}{2} * | - 33 + 12 | = \frac{1}{2} * | - 21 | = \frac{21}{2} = 10.5$ .

Like previous problem, $Δ = \frac{1}{2} | (- 1) * (- 9) + (- 3) * 8 + 5 * 9 + 3 * 0 - (0 * (- 3) + (- 9) * 5 + 8 * 3 + 9 * (- 1)) |$

Simplifying each term, we get $Δ = \frac{1}{2} | 9 - 24 + 45 + 0 - (0 - 45 + 24 - 9) |$

$Δ = \frac{1}{2} | 30 - (- 30) | = \frac{1}{2} | 30 + 30 | = 30$ .

The distance between two points in polar coordinates can be found by first converting to Cartesian coordinates: $P_{1} = (r_{1} \cos θ_{1}, r_{1} \sin θ_{1}), P_{2} = (r_{2} \cos θ_{2}, r_{2} \sin θ_{2})$

The distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $L = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .

Substituting the Cartesian coordinates, we get
$L = \sqrt{(r_{2} \cos θ_{2} - r_{1} \cos θ_{1})^{2} + (r_{2} \sin θ_{2} - r_{1} \sin θ_{1})^{2}}$

Expanding the squares, we have
$L = \sqrt{r_{2}^{2} \cos^{2} θ_{2} - 2 r_{1} r_{2} \cos θ_{1} \cos θ_{2} + r_{1}^{2} \cos^{2} θ_{1} + r_{2}^{2} \sin^{2} θ_{2} - 2 r_{1} r_{2} \sin θ_{1} \sin θ_{2} + r_{1}^{2} \sin^{2} θ_{1}}$

Grouping terms, we get
$L = \sqrt{(r_{1}^{2} (\cos^{2} θ_{1} + \sin^{2} θ_{1}) + r_{2}^{2} (\cos^{2} θ_{2} + \sin^{2} θ_{2}) - 2 r_{1} r_{2} (\cos θ_{1} \cos θ_{2} + \sin θ_{1} \sin θ_{2})}$

Using $\cos^{2} θ + \sin^{2} θ = 1$ and $\cos θ_{1} \cos θ_{2} + \sin θ_{1} \sin θ_{2} = \cos (θ_{1} - θ_{2})$ , we get $L = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} \cos (θ_{1} - θ_{2})}$ .

The distance between two points in polar coordinates is
$L = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} \cos (θ_{1} - θ_{2})}$

Substituting the given values, we get $L = \sqrt{2^{2} + 4^{2} - 2.2 .4 \cos (30^{\circ} - 120^{\circ})}$

Simplifying inside the cosine, $L = \sqrt{4 + 16 - 16. \cos (- 90^{\circ})}$ .

Since $\cos (- 90^{\circ}) = 0$ , we have $L = \sqrt{4 + 16 - 16.0} = \sqrt{20}$ .

Like previous problem $L = \sqrt{(- 3)^{2} + 7^{2} - 2. (- 3) .7 . \cos (45^{\circ} - 105^{\circ})}$ .

Simplifying, $L = \sqrt{9 + 49 - 2. (- 3) .7 . \cos (- 60^{\circ})}$ .

We know $\cos (- 60^{\circ}) = \frac{1}{2}$ , so $L = \sqrt{58 - 2. (- 3) .7 . \frac{1}{2}}$ .

Simplifying the middle term, $2. (- 3) .7 . \frac{1}{2} = - 21$ , so $L = \sqrt{58 - (- 21)} = \sqrt{58 + 21} = \sqrt{79}$ .

Hence, the distance between the points is $L = \sqrt{79}$ .

Like previous problem $L = \sqrt{a^{2} + (3 a)^{2} - 2 a .3 a . \cos (\frac{π}{2} - \frac{π}{6})}$ .

Simplifying, $L = \sqrt{a^{2} + 9 a^{2} - 6 a^{2} \cos \frac{π}{3}}$ .

Since $c o s \frac{π}{3} = \frac{1}{2}$ , we have $L = \sqrt{10 a^{2} - 3 a^{2}}$ .

$L = \sqrt{7 a^{2}} = a \sqrt{7}$ .

Let the vertices of the triangle be $P_{1} (r_{1}, θ_{1})$ , $P_{2} (r_{2}, θ_{2})$ , and $P_{3} (r_{3}, θ_{3})$ .

First, convert the polar coordinates to Cartesian coordinates: $P_{1} = (r_{1} \cos θ_{1}, r_{1} \sin θ_{1}),$
$P_{2} = (r_{2} \cos θ_{2}, r_{2} \sin θ_{2}), P_{3} = (r_{3} \cos θ_{3}, r_{3} \sin θ_{3})$ .

The area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ is $Δ = \frac{1}{2} | (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$ .

Substituting the Cartesian coordinates in terms of polar coordinates, we get $Δ = \frac{1}{2} | r_{1} \cos θ_{1} (r_{2} \sin θ_{2} - r_{3} \sin θ_{3}) + r_{2} \cos θ_{2} (r_{3} \sin θ_{3} - r_{1} \sin θ_{1}) + r_{3} \cos θ_{3} (r_{1} \sin θ_{1} - r_{2} \sin θ_{2}) |$ .

Hence, the area of the triangle is $Δ = \frac{1}{2} | r_{1} (r_{2} \sin θ_{2} - r_{3} \sin θ_{3}) \cos θ_{1} + r_{2} (r_{3} \sin θ_{3} - r_{1} \sin θ_{1}) \cos θ_{2} + r_{3} (r_{1} \sin θ_{1} - r_{2} \sin θ_{2}) \cos θ_{3} |$ .

Let the vertices be $P_{1} (1, 30^{\circ}), P_{2} (2, 60^{\circ}), P_{3} (3, 90^{\circ})$ .

Convert to Cartesian: $P_{1} = (\frac{\sqrt{3}}{2}, \frac{1}{2}), P_{2} = (1, \sqrt{3}), P_{3} = (0, 3)$

Area formula: $Δ = \frac{1}{2} | (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

Substitute: $Δ = \frac{1}{2} | \frac{\sqrt{3}}{2} (\sqrt{3} - 3) + 1 (3 - \frac{1}{2}) + 0 | = \frac{1}{2} | 4 - 3 \frac{\sqrt{3}}{2} | = 2 - 3 \frac{\sqrt{3}}{4}$ .

Let the vertices be $P_{1} (- 3, 30^{\circ}), P_{2} (5, 150^{\circ}), P_{3} (7, 210^{\circ})$ .

Convert to Cartesian coordinates: $P_{1} = (- 3 \cos 30^{\circ}, - 3 \sin 30^{\circ}) = (- 3 \frac{\sqrt{3}}{2}, - \frac{3}{2}), P_{2} = (5 \cos 150^{\circ}, 5 \sin 150^{\circ}) = (- 5 \frac{\sqrt{3}}{2}, \frac{5}{2}),$
$P_{3} = (7 \cos 210^{\circ}, 7 \sin 210^{\circ}) = (- 7 \frac{\sqrt{3}}{2}, - \frac{7}{2})$ .

Substitute the coordinates:
$Δ = \frac{1}{2} | (- 3 \frac{\sqrt{3}}{2}) (\frac{5}{2} - (- \frac{7}{2})) + (- 5 \frac{\sqrt{3}}{2}) ((- \frac{7}{2}) - (- \frac{3}{2})) + (- 7 \frac{\sqrt{3}}{2}) ((- \frac{3}{2}) - \frac{5}{2}) |$

$Δ = \frac{1}{2} | - 9 \sqrt{3} + 5 \sqrt{3} + 14 \sqrt{3} | = 5 \sqrt{3}$ .

Let the vertices be $P_{1} (- a, \frac{π}{6}), P_{2} (a, \frac{π}{2}), P_{3} (- 2 a, \frac{2 π}{3})$ .

Convert to Cartesian coordinates: $P_{1} = (- a \cos \frac{π}{6}, - a \sin \frac{π}{6}) = (- a \frac{\sqrt{3}}{2}, - \frac{a}{2}), P_{2} = (a \cos \frac{π}{2}, a \sin \frac{π}{2}) = (0, a),$
$P_{3} = (- 2 a \cos \frac{2 π}{3}, - 2 a \sin \frac{2 π}{3}) = (a, - a \frac{\sqrt{3}}{2})$ .

Substitute the coordinates:
$Δ = \frac{1}{2} | (- a \frac{\sqrt{3}}{2}) (a - (- a \frac{\sqrt{3}}{2})) + 0 (- a \frac{\sqrt{3}}{2} + \frac{a}{2}) + a (- \frac{a}{2} - a) |$

$Δ == a^{2} \frac{\sqrt{3} + 6}{4}$ .

Let the points be $P_{1} (a \cos α, a \sin α)$ and $P_{2} (a \cos β, a \sin β)$ .

The distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $L = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .

Substitute the coordinates: $L = \sqrt{(a \cos β - a \cos α)^{2} + (a \sin β - a \sin α)^{2}}$ .

Factor out $a$ in each term: $L = a \sqrt{(\cos β - \cos α)^{2} + (\sin β - \sin α)^{2}}$ .

Use the identity $(\cos β - \cos α)^{2} + (\sin β - \sin α)^{2} = 2 - 2 \cos (β - α) : L = a \sqrt{2 - 2 \cos (β - α)}$ .

Factor 2: $L = a \sqrt{2 (1 - \cos (β - α))} = a \sqrt{2} \sqrt{1 - \cos (β - α)}$ .

Let the points be $P_{1} (a t_{1}^{2}, 2 a t_{1})$ and $P_{2} (a t_{2}^{2}, 2 a t_{2})$ .

The distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $L = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .

Substitute the coordinates: $L = \sqrt{(a t_{2}^{2} - a t_{1}^{2})^{2} + (2 a t_{2} - 2 a t_{1})^{2}}$ .

Factor out $a$ from the first term and 2a from the second term: $L = \sqrt{a^{2} (t_{2}^{2} - t_{1}^{2})^{2} + (2 a (t_{2} - t_{1}))^{2}}$ .

Simplify: $(t_{2}^{2} - t_{1}^{2})^{2} = (t_{2} - t_{1})^{2} (t_{2} + t_{1})^{2}$ , $(2 a (t_{2} - t_{1}))^{2} = 4 a^{2} (t_{2} - t_{1})^{2}$ .

So $L = \sqrt{a^{2} (t_{2} - t_{1})^{2} (t_{2} + t_{1})^{2} + 4 a^{2} (t_{2} - t_{1})^{2}}$ .

Factor $a^{2} (t_{2} - t_{1})^{2} : L = \sqrt{a^{2} (t_{2} - t_{1})^{2} ((t_{2} + t_{1})^{2} + 4)}$ .

Take $a (t_{2} - t_{1})$ out of the square root: $L = a | t_{2} - t_{1} | \sqrt{(t_{2} + t_{1})^{2} + 4}$ .

The equation $x^{2} + y^{2} = a^{2}$ can be rewritten in polar coordinates. Recall that $x = r \cos θ$ and $y = r \sin θ$ . Substituting, we get:

$x^{2} + y^{2} = r^{2} \cos^{2} θ + r^{2} \sin^{2} θ = r^{2} (\cos^{2} θ + \sin^{2} θ) = r^{2}$

Hence, the equation becomes $r^{2} = a^{2}$ , or equivalently, $r = a$ .

Consider the equation $y = x \tan α$ . In polar coordinates, we have $x = r \cos θ$ and $y = r \sin θ$ .

Substituting, we get: $r \sin θ = r \cos θ \tan α$

Dividing both sides by $r \cos θ$ (assuming $r! = 0$ and $\cos θ! = 0$ ), we obtain: $\tan θ = \tan α$

Hence, in polar coordinates, the equation becomes: $θ = α + n π$ , where $n$ is any integer.

Consider the equation $x^{2} + y^{2} = 2 a x$ . In polar coordinates, $x = r \cos θ$ and $y = r \sin θ$ .

Substituting these, we get: $r^{2} \cos^{2} θ + r^{2} \sin^{2} θ = 2 a r \cos θ$

Simplifying the left side: $r^{2} (\cos^{2} θ + \sin^{2} θ) = r^{2} = 2 a r \cos θ$

Dividing both sides by $r$ (assuming $r! = 0$ ), we obtain: $r = 2 a \cos θ$

Thus, in polar coordinates, the equation becomes: $r = 2 a \cos θ$ .

Consider the equation $x^{2} - y^{2} = 2 a y$ . In polar coordinates, $x = r \cos θ$ and $y = r \sin θ$ .

Substituting these, we get: $r^{2} \cos^{2} θ - r^{2} \sin^{2} θ = 2 a r \sin θ$

Factor $r^{2}$ on the left: $r^{2} (\cos^{2} θ - \sin^{2} θ) = 2 a r \sin θ$

Divide both sides by $r$ (assuming $r! = 0$ ): $r (\cos^{2} θ - \sin^{2} θ) = 2 a \sin θ$

Hence, in polar coordinates, the equation becomes: $r = \frac{2 a \sin θ}{\cos^{2} θ - \sin^{2} θ}$ .

Consider the equation $x^{2} = y^{2} (2 a - x)$ . In polar coordinates, $x = r \cos θ$ and $y = r \sin θ$ .

Substituting these, we get: $(r \cos θ)^{2} = (r \sin θ)^{2} (2 a - r \cos θ)$

Simplifying both sides: $r^{2} \cos^{2} θ = r^{2} \sin^{2} θ (2 a - r \cos θ)$

Divide both sides by $r^{2}$ (assuming $r! = 0$ ): $\cos^{2} θ = \sin^{2} θ (2 a - r \cos θ)$

Solving for $r : r = \frac{2 a \sin^{2} θ - \cos^{2} θ}{\sin^{2} θ \cos θ}$ .

Consider the equation $(x^{2} + y^{2})^{2} = a^{2} (x^{2} - y^{2})$ .

Using $x = r \cos θ$ and $y = r \sin θ$ , we have: $(r^{2})^{2} = a^{2} r^{2} (\cos^{2} θ - \sin^{2} θ)$

Dividing both sides by $r^{2} : r^{2} = a^{2} (\cos^{2} θ - \sin^{2} θ)$ .

Given the polar equation $r = a$ , we square both sides to obtain $r^{2} = a^{2}$ .

Since $r^{2} = x^{2} + y^{2}$ , the Cartesian form is: $x^{2} + y^{2} = a^{2}$ .

Given the polar equation $θ = \tan^{- 1} m$ , taking the tangent gives: $\tan θ = m$

Since $\tan θ = \frac{y}{x}$ , we obtain the Cartesian equation: $y = m x$ .

Given the polar equation $r = a \cos θ$ , multiply both sides by $r : r^{2} = a r \cos θ$

Using $r^{2} = x^{2} + y^{2}$ and $r \cos θ = x$ , we obtain: $x^{2} + y^{2} = a x$ .

Given $r^{2} = a^{2} \sin 2 θ$

Using $x = r \cos θ, y = r \sin θ$ , and $r^{2} = x^{2} + y^{2}, s i n 2 θ = 2 \sin θ \cos θ$

Substituting: $x^{2} + y^{2} = 2 a^{2} \frac{y}{\sqrt{x^{2} + y^{2}}} . \frac{x}{\sqrt{x^{2} + y^{2}}}$

$\Rightarrow x^{2} + y^{2} = \frac{2 a^{2} x y}{x^{2} + y^{2}}$

Multiplying both sides by $x^{2} + y^{2} : (x^{2} + y^{2})^{2} = 2 a^{2} x y$ .

Given the polar equation $r^{2} \sin 2 θ = 2 a^{2}$ , use the identity $\sin 2 θ = 2 \sin θ \cos θ$ .

Since $r \sin θ = y$ and $r \cos θ = x$ , we have: $2 x y = 2 a^{2}$

Dividing both sides by $2$ gives the Cartesian equation: $x y = a^{2}$ .

Given the polar equation $\sqrt{r} \cos \frac{θ}{2} = \sqrt{a}$ , square both sides: $r \cos^{2} \frac{θ}{2} = a$

Using $c o s^{2} \frac{θ}{2} = \frac{1 + \cos θ}{2}$ , we get: $r (1 + \cos θ) = 2 a$

Since $\cos θ = \frac{x}{r}$ , this gives: $r + x = 2 a$

With $r = \sqrt{x^{2} + y^{2}}$ , the Cartesian equation is: $x^{2} + y^{2} = (2 a - x)^{2}$ .

Given the polar equation $\sqrt{r} = \sqrt{a} \sin \frac{θ}{2}$ , squaring gives: $r = a \sin^{2} \frac{θ}{2}$

Using $\sin^{2} \frac{θ}{2} = \frac{1 - \cos θ}{2}$ , we obtain: $r = a \frac{1 - \cos θ}{2}$

Substituting $\cos θ = \frac{x}{r}$ and $r^{2} = x^{2} + y^{2}$ leads to: ${(x^{2} + y^{2} + \frac{a x}{2})}^{2} = a^{2} \frac{x^{2} + y^{2}}{4}$ .

The polar equation $r (\cos 3 θ + \sin 3 θ) = 5 k \sin θ \cos θ$ can be converted to Cartesian coordinates using $x = r \cos θ, y = r \sin θ$ and $r^{2} = x^{2} + y^{2}$ .

Using the triple-angle formulas $\cos 3 θ = 4 \cos^{3} θ - 3 \cos θ$ and $\sin 3 θ = 3 \sin θ - 4 \sin^{3} θ$ , we can write:

$\cos 3 θ = \frac{4 x^{3} - 3 x r^{2}}{r^{3}}$ and $\sin 3 θ = \frac{3 y r^{2} - 4 y^{3}}{r^{3}}$

Adding these gives: $\cos 3 θ + \sin 3 θ = \frac{4 (x^{3} - y^{3}) + 3 r^{2} (y - x)}{r^{3}}$

Multiplying both sides of the original equation by $r : r (\cos 3 θ + \sin 3 θ) = \frac{4 (x^{3} - y^{3}) + 3 r^{2} (y - x)}{r^{2}}$

The right-hand side $5 k \sin θ \cos θ$ becomes $\frac{5 k x y}{r^{2}}$

Multiplying both sides by $r^{2}$ to eliminate the denominator: $4 (x^{3} - y^{3}) + 3 r^{2} (y - x) = 5 k x y$

Substituting $r^{2} = x^{2} + y^{2}$ yields the final Cartesian equation: $4 (x^{3} - y^{3}) + 3 (x^{2} + y^{2}) (y - x) = 5 k x y$ .

Distance formula is $\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}} = d^{2}$

Substituting given coordinate in the formula: $\sqrt{(a - 3)^{2} + (2 - 4)^{2}} = 8^{2}$

Simplifying gives us $a = 3 + 2 \sqrt{15}$ (rejecting negative value for distance).

Like previous problem substituting the coordinates gives us

$\sqrt{(a + r \cos θ - a)^{2} + (b + r \sin θ - b)^{2}} = r^{2}$ , which is independent of $θ$ .

Let the points be $A = (\csc^{2} θ, 0)$ , $B = (0, \sec^{2} t h e t a)$ , and $C = (1, 1)$ .

The distance formula between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is: $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Compute the distance $A B : A B = \sqrt{(0 - \csc^{2} t h e t a)^{2} + (\sec^{2} t h e t a - 0)^{2}}$

$= \sqrt{(\csc^{2} θ)^{2} + (\sec^{2} t h e t a)^{2}} = \sqrt{\csc^{4} θ + \sec^{4} θ}$

Compute the distance $B C$ :

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

Compute the distance $A C$ :

$A C = \sqrt{(1 - \csc^{2} t h e t a)^{2} + (1 - 0)^{2}} = \sqrt{(1 - \csc^{2} θ)^{2} + 1}$

Using the Pythagorean identities: $\csc^{2} θ = \frac{1}{\sin^{2} θ}$ and $\sec^{2} θ = \frac{1}{\cos^{2} θ}$

Substitute these into the distances and simplify. After simplification, we find that: $A B + B C = A C$

Therefore, the points $A, B, C$ are collinear.

Let $P = (x, y), A = (a + b, b - a)$ , and $B = (a - b, a + b)$ .

Since $P$ M is equidistant from $A$ and $B$ , we have:

$\sqrt{(x - (a + b))^{2} + (y - (b - a))^{2}} = \sqrt{(x - (a - b))^{2} + (y - (a + b))^{2}}$

Squaring both sides: $(x - (a + b))^{2} + (y - (b - a))^{2} = (x - (a - b))^{2} + (y - (a + b))^{2}$

Expanding both sides: $(x - a - b)^{2} + (y - b + a)^{2} = (x - a + b)^{2} + (y - a - b)^{2}$

Expanding each square: $(x - a - b)^{2} = x^{2} - 2 x (a + b) + (a + b)^{2} => (y - b + a)^{2} = y^{2} - 2 y (b - a) + (b - a)^{2}$

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

Adding the terms on each side:

Left: $x^{2} + y^{2} - 2 x (a + b) - 2 y (b - a) + (a + b)^{2} + (b - a)^{2}$

Right: $x^{2} + y^{2} - 2 x (a - b) - 2 y (a + b) + (a - b)^{2} + (a + b)^{2}$

Subtracting $x^{2} + y^{2} + (a + b)^{2}$ from both sides:

$- 2 x (a + b) - 2 y (b - a) + (b - a)^{2} = - 2 x (a - b) - 2 y (a + b) + (a - b)^{2}$

Simplifying $(b - a)^{2} = (a - b)^{2} : - 2 x (a + b) - 2 y (b - a) = - 2 x (a - b) - 2 y (a + b)$

Divide both sides by $- 2 : x (a + b) + y (b - a) = x (a - b) + y (a + b)$

Bringing all terms to left-hand side: $x (a + b - a + b) + y (b - a - a - b) = 0 \Rightarrow x (2 b) + y (- 2 a) = 0$

Simplifying: $2 b x - 2 a y = 0$

Dividing both sides by $b (x + y) : \frac{a - b}{a + b} = \frac{x - y}{x + y}$ .

Let the points be $A = (3, 4), B = (8, - 6),$ and $C = (13, 9)$ .

Computing the sides: $A B = \sqrt{(8 - 3)^{2} + (- 6 - 4)^{2}} = \sqrt{5^{2} + (- 10)^{2}} = \sqrt{25 + 100} = \sqrt{125}$

$B C = \sqrt{(13 - 8)^{2} + (9 - (- 6))^{2}} = \sqrt{5^{2} + 15^{2}} = \sqrt{25 + 225} = \sqrt{250}$

$A C = \sqrt{(13 - 3)^{2} + (9 - 4)^{2}} = \sqrt{10^{2} + 5^{2}} = \sqrt{100 + 25} = \sqrt{125}$

Since $A B^{2} + A C^{2} = B C^{2}$ , the $△ A B C$ is right-angled at $A$ .

Solutions are given below:

Computing the sides using the distance formula:
$A B = \sqrt{(- \sqrt{3} - 1)^{2} + (\sqrt{3} - 1)^{2}} = 2 \sqrt{2}$
$B C = \sqrt{(- 1 + \sqrt{3})^{2} + (- 1 - \sqrt{3})^{2}} = 2 \sqrt{2}$
$A C = \sqrt{(- 1 - 1)^{2} + (- 1 - 1)^{2}} = 2 \sqrt{2}$
Since $A B = B C = A C$ , the triangle is equilateral.
Let the points be $A = (0, 2), B = (7, 0),$ and $C = (2, 5)$ .
Computing the sides using the distance formula:
$A B = \sqrt{(7 - 0)^{2} + (0 - 2)^{2}} = \sqrt{49 + 4} = \sqrt{53}$
$B C = \sqrt{(2 - 7)^{2} + (5 - 0)^{2}} = \sqrt{(- 5)^{2} + 5^{2}} = \sqrt{25 + 25} = \sqrt{50}$
$A C = \sqrt{(2 - 0)^{2} + (5 - 2)^{2}} = \sqrt{4 + 9} = \sqrt{13}$
No combination satisfies the Pythagoras theorem exactly, so the triangle is not right-angled.
Since all sides are different ( $\sqrt{53}, \sqrt{50}, \sqrt{13}$ ), the triangle is scalene.
Let the points be $A = (- 2, 5), B = (7, 10),$ and $C = (3, - 4)$ .
Computing the sides using the distance formula:
$A B = \sqrt{(7 - (- 2))^{2} + (10 - 5)^{2}} = \sqrt{9^{2} + 5^{2}} = \sqrt{81 + 25} = \sqrt{106}$
$B C = \sqrt{(3 - 7)^{2} + (- 4 - 10)^{2}} = \sqrt{(- 4)^{2} + (- 14)^{2}} = \sqrt{16 + 196} = \sqrt{212}$
$A C = \sqrt{(3 - (- 2))^{2} + (- 4 - 5)^{2}} = \sqrt{5^{2} + (- 9)^{2}} = \sqrt{25 + 81} = \sqrt{106}$
$A B = A C = \sqrt{106}, B C = \sqrt{212}$
Since two sides are equal, the triangle is isosceles.
Check for a right angle using the Pythagoras theorem: $A B^{2} + A C^{2} = 106 + 106 = 212 = B C^{2}$
Hence, the triangle is right-angled isosceles at angle $A$ .

Let the point be $P = (a \cos α, a \sin α)$ and the origin be $O = (0, 0)$ .

The distance formula between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is:

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Computing the distance $O P$ :

$O P = \sqrt{(a \cos α - 0)^{2} + (a \sin α - 0)^{2}}$

Using the Pythagorean identity $\cos^{2} α + \sin^{2} α = 1$ :

$O P = \sqrt{a^{2} * 1} = \sqrt{a^{2}} = a$ , which is independent of $α$

Let the points be $A = (6, - 1)$ , $B = (1, 3)$ , and $C = (x, 8)$ .

Given $A B = B C$

$A B = \sqrt{(1 - 6)^{2} + (3 - (- 1))^{2}} = \sqrt{(- 5)^{2} + 4^{2}} = \sqrt{25 + 16} = \sqrt{41}$

$B C = \sqrt{(x - 1)^{2} + (8 - 3)^{2}} = \sqrt{(x - 1)^{2} + 5^{2}} = \sqrt{(x - 1)^{2} + 25}$

$\Rightarrow \sqrt{41} = \sqrt{(x - 1)^{2} + 25}$

Squaring both sides: $41 = (x - 1)^{2} + 25 \Rightarrow (x - 1)^{2} = 16$

$\Rightarrow x - 1 = 4$ or $x - 1 = - 4 \Rightarrow x = 5$ or $x = - 3$ .

Let the points be $A = (1, 5)$ , $B = (2, 4)$ , and $C = (3, 3)$ .

$A B = \sqrt{(2 - 1)^{2} + (4 - 5)^{2}} = \sqrt{1^{2} + (- 1)^{2}} = \sqrt{1 + 1} = \sqrt{2}$

$B C = \sqrt{(3 - 2)^{2} + (3 - 4)^{2}} = \sqrt{1^{2} + (- 1)^{2}} = \sqrt{1 + 1} = \sqrt{2}$

$A C = \sqrt{(3 - 1)^{2} + (3 - 5)^{2}} = \sqrt{2^{2} + (- 2)^{2}} = \sqrt{4 + 4} = \sqrt{8}$

Check if $A B + B C = A C : A B + B C = \sqrt{2} + \sqrt{2} = \sqrt{2}$

$A C = \sqrt{8} = 2 \sqrt{2}$

Since $A B + B C = A C$ , the points $A$ , $B$ , and $C$ are collinear.

Let the points be $A = (2 a, 4 a)$ , $B = (2 a, 6 a)$ , $C = (2 a + \sqrt{3} a, 5 a)$ .

$A B = \sqrt{(2 a - 2 a)^{2} + (6 a - 4 a)^{2}} = \sqrt{0 + (2 a)^{2}} = \sqrt{4 a^{2}} = 2 a$

$B C = \sqrt{(2 a + \sqrt{3} a - 2 a)^{2} + (5 a - 6 a)^{2}} = \sqrt{(\sqrt{3} a)^{2} + (- a)^{2}} = \sqrt{3 a^{2} + a^{2}} = \sqrt{4 a^{2}} = 2 a$

$A C = \sqrt{(2 a + \sqrt{3} a - 2 a)^{2} + (5 a - 4 a)^{2}} = \sqrt{(\sqrt{3} a)^{2} + (a)^{2}} = \sqrt{3 a^{2} + a^{2}} = \sqrt{4 a^{2}} = 2 a$

Since $A B = B C = A C = 2 a$ , all sides are equal.

Let the points be $O = (0, 0)$ , $A = (a, b)$ , and $B = (c, d)$ .

$O A = \sqrt{(a - 0)^{2} + (b - 0)^{2}} = \sqrt{a^{2} + b^{2}}$

$O B = \sqrt{(c - 0)^{2} + (d - 0)^{2}} = \sqrt{c^{2} + d^{2}}$

$A B = \sqrt{(c - a)^{2} + (d - b)^{2}} = \sqrt{(c - a)^{2} + (d - b)^{2}}$

By the cosine law, for triangle $O A B$ with angle $θ$ at $O$ :

$\cos θ = \frac{O A^{2} + O B^{2} - A B^{2}}{2. O A . O B}$

Substituting the distances:

$\cos θ = \frac{(a^{2} + b^{2}) + (c^{2} + d^{2}) - ((c - a)^{2} + (d - b)^{2})}{2 \sqrt{a^{2} + b^{2}} \sqrt{c^{2} + d^{2}}}$

Expanding $A B^{2} = (c - a)^{2} + (d - b)^{2} = c^{2} - 2 a c + a^{2} + d^{2} - 2 b d + b^{2} = a^{2} + b^{2} + c^{2} + d^{2} - 2 (a c + b d)$

Substituting: $\cos θ = \frac{(a^{2} + b^{2}) + (c^{2} + d^{2}) - (a^{2} + b^{2} + c^{2} + d^{2} - 2 (a c + b d))}{2 \sqrt{a^{2} + b^{2}} \sqrt{c^{2} + d^{2}}}$

$\Rightarrow \cos θ = \frac{2 (a c + b d)}{2 \sqrt{a^{2} + b^{2}} \sqrt{c^{2} + d^{2}}}$

$\Rightarrow \cos θ = \frac{a c + b d}{\sqrt{a^{2} + b^{2}} \sqrt{c^{2} + d^{2}}}$ .

Let the vertices be $A = (- 2, - 3)$ , $B = (- 1, 0)$ , and $C = (7, - 6)$ .

Let the circumcenter be $O = (x, y)$ , which is equidistant from all three vertices:

$\sqrt{(x + 2)^{2} + (y + 3)^{2}} = \sqrt{(x + 1)^{2} + y^{2}}$

Squaring both sides: $(x + 2)^{2} + (y + 3)^{2} = (x + 1)^{2} + y^{2}$

$\Rightarrow x^{2} + 4 x + 4 + y^{2} + 6 y + 9 = x^{2} + 2 x + 1 + y^{2} \Rightarrow 2 x + 6 y + 12 = 0 \Rightarrow x + 3 y + 6 = 0$

Also, > $\sqrt{(x + 1)^{2} + y^{2}} = \sqrt{(x - 7)^{2} + (y + 6)^{2}}$

Squaring both sides: $(x + 1)^{2} + y^{2} = (x - 7)^{2} + (y + 6)^{2}$

$\Rightarrow x^{2} + 2 x + 1 + y^{2} = x^{2} - 14 x + 49 + y^{2} + 12 y + 36$

$\Rightarrow 16 x - 12 y - 84 = 0 => 4 x - 3 y - 21 = 0$

Solving the system: From $x + 3 y + 6 = 0$ , we get $x = - 3 y - 6$

Figure 7.1.

Substitute into $4 x - 3 y - 21 = 0 : 4 (- 3 y - 6) - 3 y - 21 = 0 \Rightarrow - 15 y - 45 = 0 \Rightarrow y = - 3$

Then $x = - 3 (- 3) - 6 =$ 3

Thus, the circumcenter is $O = (3, - 3)$

The circumradius is: $R = \sqrt{(3 - (- 2))^{2} + (- 3 - (- 3))^{2}} = \sqrt{5^{2} + 0^{2}} = 5$

The diagram is given below:

Figure 7.2.

Let $P Q = Q R = R P = r$ and $∠ P Q X = θ$ then $∠ R Q X = 60^{\circ} + θ$ .

Given $P L = a, R N = 1$ >.

We take $Q$ as the pole and $Q X$ as the initial line. Polar coordinates of $P$ and $R$ will be $(r, θ)$ and $(r, θ + 60^{\circ})$ , respectively.

Now $r \sin θ = y$ , coordinate of $P = a$ and $r s i n (60^{\circ} + θ) = y$ , coordinate of $R = 1$

$\Rightarrow r (\sin 60^{\circ} \cos θ + \cos 60^{\circ} \sin θ) = 1$

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

$\Rightarrow r = \frac{2}{\sqrt{3}} \sqrt{a^{2} - a + 1}$ .

Let $A = (3, 4)$ and $C = (1, - 1)$ be opposite vertices of the square. Let $B = (x, y)$ and $D = (x^{'}, y^{'})$ be the other two vertices.

The center of the square is the midpoint of $A C : O = (\frac{3 + 1}{2}, \frac{4 + (- 1)}{2}) = (2, \frac{3}{2})$

In a square, the distances from the center to all vertices are equal: $(x - 2)^{2} + {(y - \frac{3}{2})}^{2} = (3 - 2)^{2} + {(4 - \frac{3}{2})}^{2} = \frac{29}{4}$

Also, $B$ and $D$ lie on a line through $O$ that is perpendicular to $A C$ . Using the fact that diagonals of a square are equal and perpendicular, we solve:

$(x - 2)^{2} + {(- \frac{2}{5} x + \frac{23}{10} - \frac{3}{2})}^{2} = \frac{29}{4}$

Solving this quadratic gives: $x = \frac{9}{2}, y = \frac{1}{2}$ and $x = - \frac{1}{2}, y = \frac{5}{2}$

Hence, the remaining vertices are: $B = (\frac{9}{2}, \frac{1}{2}), D = (- \frac{1}{2}, \frac{5}{2})$ .

Given $A = (- 4, 0)$ and $B = (- 1, 4)$ , points $C$ and $D$ are symmetric about the $y$ -axis: $C = (4, 0)$ and $D = (1, 4)$

The trapezium is $A B C D$ , with vertices in order $A, B, D, C$ .

$A B = \sqrt{(- 1 - (- 4))^{2} + (4 - 0)^{2}} = \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = 5$

$B C = \sqrt{(1 - (- 1))^{2} + (4 - 4)^{2}} = \sqrt{2^{2} + 0^{2}} = 2$

$C D = \sqrt{(1 - 4)^{2} + (4 - 0)^{2}} = \sqrt{(- 3)^{2} + 4^{2}} = \sqrt{9 + 16} = 5$

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

Perimeter of trapezium: $P = A B + B C + C D + D A = 5 + 2 + 5 + 8 = 20$ .

Given $A = (4, - 1)$ , let $B$ be the reflection of $A$ across $y = x$ .

Reflection across $y = x$ swaps the coordinates: if $A = (x, y)$ , then $B = (y, x)$ . Thus $B = (- 1, 4)$ .

Distance $A B : A B = \sqrt{(- 1 - 4)^{2} + (4 - (- 1))^{2}} = 5 \sqrt{2}$ .Given $A = (4, - 1)$ , let $B$ be the reflection of $A$ across $y = x$ .

Reflection across $y = x$ swaps the coordinates: if $A = (x, y)$ , then $B = (y, x)$ . Thus $B = (- 1, 4)$ .

Distance $A B : A B = \sqrt{(- 1 - 4)^{2} + (4 - (- 1))^{2}} = 5 \sqrt{2}$ .

Given $A = (2, 0)$ and $B = (2 + \sqrt{3}, 1)$ . Rotate $B$ about $A$ by $15^{\circ}$ anticlockwise.

Rotation formula about $A$ :

$x^{'} = A_{x} + (B_{x} - A_{x}) \cos 15^{\circ} - (B_{y} - A_{y}) \sin 15^{\circ}, y^{'} = A_{y} + (B_{x} - A_{x}) \sin 15^{\circ} + (B_{y} - A_{y}) \cos 15^{\circ}$

$(B_{x} - A_{x}, B_{y} - A_{y}) = (\sqrt{(3}, 1)$ , $c o s 15^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$ , $s i n 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}$

$x^{'} = 2 + \sqrt{3} \frac{\sqrt{6} + \sqrt{2}}{4} - 1 \frac{\sqrt{6} - \sqrt{2}}{4} = \frac{11 + \sqrt{2}}{4}, y^{'} = 0 + \sqrt{3} \frac{\sqrt{6} - \sqrt{2}}{4} + 1 \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{3 + \sqrt{2}}{4}$

Thus, $C = (\frac{11 + \sqrt{2}}{4}, \frac{3 + \sqrt{2}}{4})$ .

Given point $P = (1, - 2)$ . Reflecting $P$ in the $x$ -axis: $P^{'} = (1, 2)$ (the $y$ -coordinate changes sign)

Translate $P^{'}$ parallel to the positive $x$ -axis by $3$ units: $P^{″} = (1 + 3, 2) = (4, 2)$ .

Given $A = (3, 0)$ and $B = (5, 2)$ .

Segment from $A$ to $B : (B_{x} - A_{x}, B_{y} - A_{y}) = (5 - 3, 2 - 0) = (2, 2)$

Rotating this segment about $A$ by $45^{\circ}$ anticlockwise:

$x_{C} = A_{x} + (B_{x} - A_{x}) \cos 45^{\circ} - (B_{y} - A_{y}) \sin 45^{\circ}, y_{C} = A_{y} + (B_{x} - A_{x}) \sin 45^{\circ} + (B_{y} - A_{y}) \cos 45^{\circ}$

Since $\cos 45^{\circ} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$ , $x_{C} = 3 + 2 \frac{1}{\sqrt{2}} - 2 \frac{1}{\sqrt{2}} = 3 + 0 = 3$

$y_{C} = 0 + 2 \frac{1}{\sqrt{2}} + 2 \frac{1}{\sqrt{2}} = 0 + 2 \frac{1}{\sqrt{2}} + 2 \frac{1}{\sqrt{2}} = 2 \sqrt{2}$

So $C = (3, 2 \sqrt{2})$ . Reflecting $C$ in the $y$ -axis: $D = (- x_{C}, y_{C}) = (- 3, 2 \sqrt{2})$ .

Let $A (5, - 2)$ and $B (9, 6)$ .

For a point dividing the line segment joining $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ in the ratio $m : n$ :

Internal division formula is $(\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + 1})$ .

External division formula is $(\frac{m x_{2} - n x_{1}}{m - n}, \frac{m y_{2} - n y_{1}}{m - n})$ .

Here $m = 3$ and $n = 1$ .

Internal division point is $(\frac{3.9 + 1.5}{4}, \frac{3.6 + 1. (- 2)}{4}) = (8, 4)$ .

External division point is $(\frac{3.9 - 1.5}{2}, \frac{3.6 - 1. (- 2)}{2}) = (11, 10)$ .

The roots of $x^{2} + 4 x + 3 = 0$ are $- 3$ and $- 1$ . Since $x_{B} < x_{C}$ , we have $x_{B} = - 3$ and $x_{C} = - 1$ .

The roots of $x^{2} - x - 6 = 0$ are $3$ and $- 2$ . Since $y_{B} > y_{C}$ , we have $y_{B} = 3$ and $y_{C} = - 2$ .

So the coordinates are $B (- 3, 3)$ , $C (- 1, - 2)$ , and $A (3, - 5)$ .

Now, $B C = \sqrt{(- 1 + 3)^{2} + (- 2 - 3)^{2}} = \sqrt{4 + 25} = \sqrt{29}$ ,

$C A = \sqrt{(3 + 1)^{2} + (- 5 + 2)^{2}} = \sqrt{16 + 9} = 5$ ,

$A B = \sqrt{(3 + 3)^{2} + (- 5 - 3)^{2}} = \sqrt{36 + 64} = 10$ .

Let $a = B C = \sqrt{29}$ , $b = C A = 5$ , and $c = A B = 10$ .

The length of the internal angle bisector at $A$ is $l_{a} = \frac{\sqrt{b c ((b + c)^{2} - a^{2})}}{b + c}$ .

Substituting values, $l_{a} = \frac{\sqrt{5.10 (15^{2} - 29)}}{15} = \frac{\sqrt{50.196}}{15} = \frac{14 \sqrt{2}}{3}$ .

Let the points be $A (4, 3)$ and $B (6, 3)$ . Let $P (2, y)$ divide $A B$ in the ratio $m : n$ .

Using the section formula for the $x$ -coordinate, $2 = \frac{6 m + 4 n}{m + n}$ .

So, $2 m + 2 n = 6 m + 4 n \Rightarrow 4 m + 2 n = 0 \Rightarrow m : n = - 1 : 2$ .

Hence, the point divides the line segment externally in the ratio $1 : 2$ .

Now using the $y$ -coordinate formula, $y = \frac{3 m + 3 n}{m + n} \Rightarrow y = 3$ .

The given points are $A (- 2, 3)$ , $B (8, 9)$ , $C (0, 4)$ and $D (3, 0)$ .

Let the lines $A B$ and $C D$ intersect at point $P$ .

Equation of $A B$ in parametric form is $x = - 2 + 10 t$ , $y = 3 + 6 t$ .

Equation of $C D$ in parametric form is $x = 3 s$ , $y = 4 - 4 s$ .

At the point of intersection, $- 2 + 10 t = 3 s \Rightarrow 3 + 6 t = 4 - 4 s$ .

From the first equation, $s = \frac{- 2 + 10 t}{3}$ .

Substituting in the second equation, $3 + 6 t = 4 - 4 \frac{- 2 + 10 t}{3} \Rightarrow t = \frac{11}{58}$ .

Therefore, $A P : P B = t : (1 - t) = (11 / 58) : (47 / 58) = 11 : 47$ .

Let $A_{1} (x_{1}, y_{1}), A_{2} (x_{2}, y_{2}), A_{3} (x_{3}, y_{3}), \dots, A_{n} (x_{n}, y_{n})$ be $n$ points.

$A_{1} A_{2}$ is bisected at $G_{1}$ , so $G_{1} = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

$G_{1} A_{3}$ is divided at $G_{2}$ in the ratio $1 : 2$ , hence $G_{2} = (\frac{1. x_{3} + 2. x . G_{1}}{3}, \frac{1. y_{3} + 2. y . G_{1}}{3})$ .

Substituting the coordinates of $G_{1}$ , $G_{2} = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ .

Now $G_{2} A_{4}$ is divided at $G_{3}$ in the ratio $1 : 3$ , so $G_{3} = (\frac{1. x_{4} + 3. x . G_{2}}{4}, \frac{1. y_{4} + 3. y . G_{2}}{4})$ .

Substituting the coordinates of $G_{2}$ , $G_{3} = (\frac{x_{1} + x_{2} + x_{3} + x_{4}}{4}, \frac{y_{1} + y_{2} + y_{3} + y_{4}}{4})$ .

Proceeding similarly, after dividing $G_{n - 2} A_{n}$ at $G_{n - 1}$ in the ratio $1 : (n - 1)$ , we get

$G_{n - 1} = (\frac{x_{1} + x_{2} + \dots + x_{n}}{n}, \frac{y_{1} + y_{2} + \dots + y_{n}}{n})$ .

Let the straight line be $a x + b y + c = 0$ and let it intersect the line joining $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ at the point $P$ .

Suppose $P$ divides $A B$ in the ratio $m : n$ .

Then, by the section formula, the coordinates of $P$ are $x = \frac{m x_{2} + n x_{1}}{m + n}$ and $y = \frac{m y_{2} + n y_{1}}{m + n}$

Since $P$ lies on the line $a x + b y + c = 0$ , we have $a \frac{m x_{2} + n x_{1}}{m + n} + b \frac{m y_{2} + n y_{1}}{m + n} + c = 0$ .

Multiplying by $(m + n), m (a x_{2} + b y_{2} + c) + n (a x_{1} + b y_{1} + c) = 0$ .

Hence, $m (a x_{2} + b y_{2} + c) = - n (a x_{1} + b y_{1} + c)$ .

Therefore, $m : n = - (a x_{1} + b y_{1} + c) : (a x_{2} + b y_{2} + c)$ .

The negative sign means that the quantities $a x_{1} + b y_{1} + c$ and $a x_{2} + b y_{2} + c$ are of opposite signs.

Geometrically, this implies that the points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ lie on opposite sides of the line $a x + b y + c = 0$ .

Let the line $L$ intersect $B C$ at $P$ , $C A$ at $Q$ , and $A B$ at $R$ .

Equation of the line $L$ may be written as $a x + b y + c = 0$ .

Since $P$ lies on $B C$ , by the result for division of a line by a straight line, $\frac{B P}{P C} = - \frac{a x_{2} + b y_{2} + c}{a x_{3} + b y_{3} + c}$ .

Similarly, since $Q$ lies on $C A, \frac{C Q}{Q A} = - \frac{a x_{3} + b y_{3} + c}{a x_{1} + b y_{1} + c}$ .

And since $R$ lies on $A B, \frac{A R}{R B} = - \frac{a x_{1} + b y_{1} + c}{a x_{2} + b y_{2} + c}$ .

Multiplying the three ratios, we get $\frac{B P}{P C} . \frac{C Q}{Q A} . \frac{A R}{R B}$
$= - \frac{(a x_{2} + b y_{2} + c) (a x_{3} + b y_{3} + c) (a x_{1} + b y_{1} + c)}{(a x_{3} + b y_{3} + c) (a x_{1} + b y_{1} + c) (a x_{2} + b y_{2} + c)} = - 1$ .

The vertices of the triangle are $A = (x_{1}, x_{1} \tan α)$ , $B = (x_{2}, x_{2} \tan β)$ , $C = (x_{3}, x_{3} \tan γ)$ .

Since the circumcenter is at the origin, we have $x_{1}^{2} + (x_{1} \tan α)^{2} = x_{2}^{2} + (x_{2} \tan β)^{2} = x_{3}^{2} + (x_{3} \tan γ)^{2}$ .

Hence $x_{1} \sec α = x_{2} \sec β = x_{3} \sec γ = R$ .

The coordinates of the orthocenter $H (\overset{―}{x}, \overset{―}{y})$ are $\overset{―}{x} = x_{1} + x_{2} + x_{3}$ , $\overset{―}{y} = x_{1} \tan α + x_{2} \tan β + x_{3} \tan γ$ .

Substituting $x_{i} = R \cos θ_{i}$ , we get $\overset{―}{x} = R (\cos α + \cos β + \cos γ)$ , $\overset{―}{y} = R (\cos α \tan α + \cos β \tan β + \cos γ \tan γ) = R (\sin α + \sin β + \sin γ)$ .

Therefore, $\frac{\overset{―}{y}}{\overset{―}{x}} = \frac{\sin α + \sin β + \sin γ}{\cos α + \cos β + \cos γ}$ .

The vertices of the triangle are $A (α, \frac{1}{α})$ , $B (β, \frac{1}{β})$ , and $C (γ, \frac{1}{γ})$ , where $α$ , $β$ , and $γ$ are the roots of $x^{3} - 3 p x^{2} + 3 q x - 1 = 0$ .

The centroid of the triangle is $G = (\frac{α + β + γ}{3}, \frac{\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}}{3})$ .

Since $α + β + γ = 3 p$ and $\frac{1}{α} + \frac{1}{β} + \frac{1}{γ} = \frac{β γ + γ α + α β}{α β γ} = 3 q$ , we have

$G = (p, q)$ .

Let the points be $A (a t^{2}, 2 a t)$ , $B (\frac{a}{t^{2}}, - \frac{2 a}{t})$ , and $C (a, 0)$ .

The distance between $A$ and $C$ is $A C = \sqrt{{(a t^{2} - a)}^{2} + (2 a t - 0)^{2}} = a (t^{2} + 1)$

The distance between $B$ and $C$ is $B C = \sqrt{{(\frac{a}{t^{2}} - a)}^{2} + {(- \frac{2 a}{t} - 0)}^{2}} = a \frac{t^{2} + 1}{t^{2}}$

Then $\frac{1}{A C} + \frac{1}{B C} = \frac{1}{a (t^{2} + 1)} + \frac{t^{2}}{a (t^{2} + 1)} = \frac{1}{a}$ .

Hence, $\frac{1}{A C} + \frac{1}{B C}$ is independent of $t$ .

Let $A = (0, 0)$ , $B = (3, \sqrt{3})$ , and $C = (x, y)$ . $A B = 2 \sqrt{3}$ , so $A C = B C = 2 \sqrt{3}$ .

From distances: $x^{2} + y^{2} = 12$ , $(x - 3)^{2} + {(y - \sqrt{3})}^{2} = 12$ .

Simplifying gives $y = 2 \sqrt{3} - \sqrt{3} x$ .

Substituting in $x^{2} + y^{2} = 12 : x^{2} + {(2 \sqrt{3} - \sqrt{3} x)}^{2} = 12 \Rightarrow 4 x (x - 3) = 0 \Rightarrow x = 0$ or $3$ .

Then $y = 2 \sqrt{3}$ or $y = - \sqrt{3}$ .

Hence, $C = (0, 2 \sqrt{3})$ or $C = (3, - \sqrt{3})$ .

Let the vertices of the triangle be $A (- 2, 3)$ , $B (2, - 1)$ , $C (4, 0)$ .

The circumcenter $(x, y)$ is given by

$x = \frac{(x_{1}^{2} + y_{1}^{2}) (y_{2} - y_{3}) + (x_{2}^{2} + y_{2}^{2}) (y_{3} - y_{1}) + (x_{3}^{2} + y_{3}^{2}) (y_{1} - y_{2})}{2 (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}))}$

$y = \frac{(x_{1}^{2} + y_{1}^{2}) (x_{3} - x_{2}) + (x_{2}^{2} + y_{2}^{2}) (x_{1} - x_{3}) + (x_{3}^{2} + y_{3}^{2}) (x_{2} - x_{1})}{2 (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}))}$

Denominator: $D = 2 ((- 2) (- 1 - 0) + 2 (0 - 3) + 4 (3 - (- 1))) = 24$

$x = ((- 2^{2} + 3^{2}) (- 1 - 0) + (2^{2} + (- 1)^{2}) (0 - 3) + (4^{2} + 0^{2}) (3 - (- 1))) / 24 = 1.5$

$y = ((- 2^{2} + 3^{2}) (4 - 2) + (2^{2} + (- 1)^{2}) (- 2 - 4) + (4^{2} + 0^{2}) (2 - (- 2))) / 24 = 2.5$

Hence the circumcenter is $O (1.5, 2.5)$ .

The circumradius is the distance from $O$ to any vertex, say $A$ :

$R = \sqrt{(1.5 - (- 2))^{2} + (2.5 - 3)^{2}} = \sqrt{{3.5}^{2} + (- 0.5)^{2}} = \sqrt{12.25 + 0.25} = \sqrt{12.5} = \frac{5}{\sqrt{2}}$ .

Given $A = (1, 1)$ , $B = (4, 5)$ , and $C = (6, 13)$ .

$A B = \sqrt{(4 - 1)^{2} + (5 - 1)^{2}} = \sqrt{9 + 16} = 5$

$A C = \sqrt{(6 - 1)^{2} + (13 - 1)^{2}} = \sqrt{25 + 144} = 13$

$B C = \sqrt{(6 - 4)^{2} + (13 - 5)^{2}} = \sqrt{4 + 64} = \sqrt{68}$

Using cosine law at angle $A : \cos A = \frac{A B^{2} + A C^{2} - B C^{2}}{2 A B . A C}$

$\cos A = \frac{25 + 169 - 68}{2.5 .13} = \frac{63}{65}$ .

The given points are $(3, \frac{π}{4})$ and $(7, \frac{5 π}{4})$ .

Distance between two points in polar coordinates is: $d^{2} = r_{1}^{2} + r_{2}^{2} - 2. r_{1} . r_{2} . \cos (θ_{2} - θ_{1})$

Here, $r_{1} = 3$ , $r_{2} = 7$ and $θ_{2} - θ_{1} = \frac{5 π}{4} - \frac{π}{4} = π$ . Now $\cos π = - 1$

Substituting: $d^{2} = 3^{2} + 7^{2} - 2.3 .7 . (- 1)$ $\Rightarrow d^{2} = 9 + 49 + 42 = 100 \Rightarrow d = \sqrt{100} = 10$ .

Given $A = (2, 4)$ and $B = (2, 6)$ . Length $A B = \sqrt{(2 - 2)^{2} + (6 - 4)^{2}} = 2$ .

Midpoint of $A B : M = (\frac{2 + 2}{2}, \frac{4 + 6}{2}) = (2, 5)$

Since $A B$ is vertical, the equilateral triangle lies horizontally. Height of an equilateral triangle with side $2$ is: $h = \sqrt{3}$

The origin $(0, 0)$ lies to the left of line $x = 2$ , so the vertex opposite to the origin lies to the right.

Hence, $P = (2 + \sqrt{3}, 5)$ .

Given points in polar coordinates: $A = (2, 45^{\circ})$ , $B = (\sqrt{2}, 90^{\circ})$ , and $C = (- 2, 135^{\circ})$ .

Converting to cartesian coordinates using $x = r \cos θ$ and $y = r \sin θ$

$A = (2 \cos 45^{\circ}, 2 \sin 45^{\circ}) = (\sqrt{2}, \sqrt{2})$ , $B = (\sqrt{2} \cos 90^{\circ}, \sqrt{2} \sin 90^{\circ}) = (0, \sqrt{2})$ , and $C = (- 2 \cos 135^{\circ}, - 2 \sin 135^{\circ}) = (\sqrt{2}, - \sqrt{2})$

Finding squares of side lengths using distance formula: $A B^{2} = {(0 - \sqrt{2})}^{2} + {(\sqrt{2} - \sqrt{2})}^{2} = 2$

$A C^{2} = {(\sqrt{2} - \sqrt{2})}^{2} + {(- \sqrt{2} - \sqrt{2})}^{2} = 8$

$B C^{2} = {(\sqrt{2} - 0)}^{2} + {(- \sqrt{2} - \sqrt{2})}^{2} = 10$

$A B^{2} + A C^{2} = 2 + 8 = 10 = B C^{2}$ . And hence, the triangle is a right-angled triangle.

Let $A = (2, 4)$ and $B = (6, 8)$ .

Internal division in the ratio $1 : 3 : x = \frac{1.6 + 3.2}{1 = 3} = \frac{12}{4} = 3$ , and $y = \frac{1.8 + 3.4}{1 + 3} = \frac{20}{4} = 5$

So, the point of internal division is $(3, 5)$ .

External division in the ratio $1 : 3 : x = \frac{1.6 - 3.2}{1 - 3} = \frac{0}{- 2} = 0$ , and $y = \frac{1.8 - 3.4}{1 - 3} = \frac{- 4}{- 2} (- 4) / (- 2) = 2$

So, the point of external division is $(0, 2)$ .

The solutions are given below:

Let $A = (1, - 2)$ and $B = (- 3, 4)$ . The trisection points divide the line internally in the ratios $1 : 2$ and $2 : 1$ .
First point (ratio $1 : 2$ ): $x = \frac{1. - 3 + 2.1}{1 + 2} = \frac{- 3 + 2}{3} = - \frac{1}{3}$ , and $y = \frac{1.4 + 2. - 2}{1 + 2} = 0$
So, the first trisection point is $(- 1 / 3, 0)$ .
Second point (ratio $2 : 1$ ): $x = \frac{2. - 3 + 1.1}{2 + 1} = \frac{- 6 + 1}{3} = - \frac{5}{3}$ , and $y = \frac{2.4 + 1. - 2}{2 + 1} = \frac{8 - 2}{3} = 2$
So, the second trisection point is $(- 5 / 3, 2)$ .
Let $A = (2, 3)$ and $B = (6, 5)$ . The trisection points divide the line internally in the ratios $1 : 2$ and $2 : 1$ .
First point (ratio $1 : 2$ ): $x = \frac{1.6 + 2.2}{1 + 2} = \frac{10}{3}$ , and $y = \frac{1.5 + 2.3}{1 + 2} = \frac{11}{3}$
So, the first trisection point is $(10 / 3, 11 / 3)$ .
Second point (ratio $2 : 1$ ): $x = \frac{2.6 + 1.2}{2 + 1} = \frac{14}{3}$ , and $y = \frac{2.5 + 1.3}{2 + 1} = \frac{13}{3}$
So, the second trisection point is $(14 / 3, 13 / 3)$ .

Let $A = (1, 1)$ and $B = (2, - 3)$ . Point $D$ lies on $A B$ produced such that $A D = 3 A B$ .

The ratio of division for external point $D$ is $A : B = 3 : - 1$ .

Coordinates of $D$ using external division: $x_{D} = \frac{3.2 - 1.1}{3 - 1} = \frac{5}{2}$ , and $y_{D} = \frac{3. - 3 - 1.1}{3 - 1} = - 5$

So, the coordinates of $D$ are $(5 / 2, - 5)$ .

Let the points be $A = (3, 4)$ and $B = (k, 7)$ .

Midpoint $M = (x, y)$ is given by: $x = \frac{3 + k}{2}$ , and $y = \frac{4 + 7}{2} = \frac{11}{2}$

The line passing through $M$ satisfies: $2 x + 2 y + 1 = 0$

Substituting $x$ and $y : 2. \frac{3 + k}{2} + 2. \frac{11}{2} + 1 = 0$

$\Rightarrow (3 + k) + 11 + 1 = 0 \Rightarrow k + 15 = 0 \Rightarrow k = - 15$ .

Let one end of the diameter be $A = (2, 3)$ and the center be $O = (- 2, 5)$ . Let the other end be $B = (x, y)$ .

The midpoint of $A B$ is the center $O$ , so $O_{x} = \frac{A_{x} + B_{x}}{2}$ , $O_{y} = \frac{A_{y} + B_{y}}{2}$

Substituting the known values: $- 2 = \frac{2 + x}{2} \Rightarrow 2 + x = - 4 \Rightarrow x = - 6$

$5 = \frac{3 + y}{2} \Rightarrow 3 + y = 10 \Rightarrow y = 7$

Thus, the coordinates of the other end of the diameter are: $B = (- 6, 7)$ .

Let the vertices be $A = (- 1, 3)$ , $B = (1, - 1)$ , $C = (5, 1)$ .

Median from $A$ to midpoint of $B C$ : Midpoint of $B C$ : $M_{a} = (\frac{1 + 5}{2}, \frac{- 1 + 1}{2}) = (\frac{6}{2}, \frac{0}{2}) = (3, 0)$

Length of median $A M_{a} : A M_{a} = \sqrt{(3 - (- 1))^{2} + (0 - 3)^{2}} = \sqrt{4^{2} + (- 3)^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$

Median from $D$ to midpoint of $A C$ : Midpoint of $A C : M_{b} = (\frac{- 1 + 5}{2}, \frac{3 + 1}{2}) = (\frac{4}{2}, \frac{4}{2}) = (2, 2)$

Length of median $B M_{b} : B M_{b} = \sqrt{(2 - 1)^{2} + (2 - (- 1))^{2}} = \sqrt{1^{2} + 3^{2}} = \sqrt{1 + 9} = \sqrt{10}$

Median from $C$ to midpoint of $A B$ : Midpoint of $A B : M_{c} = (\frac{- 1 + 1}{2}, \frac{3 + (- 1)}{2}) = (0, 1)$

Length of median $C M_{c} : C M_{c} = \sqrt{(0 - 5)^{2} + (1 - 1)^{2}} = \sqrt{(- 5)^{2} + 0^{2}} = \sqrt{25} = 5$ .

Let $A = (2, 5)$ , $C = (- 1, 2)$ , and $B = (x, y)$ . Point $C$ divides $A B$ in the ratio $3 : 4$ .

Using section formula for internal division:

$C_{x} = \frac{4.2 + 3. x}{3 + 4} \Rightarrow - 1 = \frac{8 + 3. x}{7} \Rightarrow 8 + 3. x = - 7 \Rightarrow 3. x = - 15 \Rightarrow x = - 5$

$C_{y} = \frac{4.5 + 3. y}{3 + 4} \Rightarrow 2 = \frac{20 + 3 y}{7} \Rightarrow 20 + 3. y = 14 \Rightarrow 3. y = - 6 \Rightarrow y = - 2$ .

Let $A = (3, 4)$ , $B = (7, 7)$ , $C = (x, y)$ with $B$ between $A$ and $C$ .

Distance $A B = \sqrt{(7 - 3)^{2} + (7 - 4)^{2}} = 5$ . Distance $A C = 10 \Rightarrow B C = A C - A B = 5$

Since $B$ divides $A C$ internally in the ratio $A B : B C = 5 : 5 = 1 : 1$ .

Use section formula for internal division: $B_{x} = \frac{x + 3}{2} \Rightarrow 7 = \frac{x + 3}{2} \Rightarrow x = 11$ , and $B_{y} = \frac{y + 4}{2} \Rightarrow 7 = \frac{y + 4}{2} \Rightarrow y = 10$ .

Let $A = (2, - 2)$ , $B = (- 4, 1)$ , and $P = (- 8, 3)$ divides $A B$ in the ratio $k : 1$ .

Using section formula: $P_{x} = \frac{k . (- 4) + 1.2}{k + 1} \Rightarrow - 8 = \frac{- 4 k + 2}{k + 1}$

Multiply both sides by $(k + 1) : - 8. (k + 1) = - 4 k + 2 \Rightarrow - 8 k - 8 = - 4 k + 2 \Rightarrow - 8 k + 4 k = 2 + 8 \Rightarrow - 4 k = 10 \Rightarrow k = - 5 / 2$ .

Let $A = (2, - 3)$ , $B = (5, 6)$ , and $P = (x, 0)$ be the point where the line meets the $x$ -axis.

Suppose $P$ divides $A B$ in the ratio $k : 1$ . Using section formula for $y$ -coordinate:

$0 = \frac{k .6 + 1. (- 3)}{k + 1} \Rightarrow 0 = \frac{6 k - 3}{k + 1} \Rightarrow 6 k - 3 = 0 \Rightarrow 6 k = 3 \Rightarrow k = 1 / 2$ .

Let $C = (- 1, 2)$ , $D = (4, - 5)$ , and $P$ be the intersection of line $A B$ with $C D$ . Suppose $P$ divides $C D$ in the ratio $m : n$ .

$P_{x} = \frac{- n + 4 m}{m + n}$ , $P_{y} = \frac{2 n - 5 n}{m + n}$

Line $A B : X = 15 t, Y = - 1 + 3 t$

Equating coordinates: $15 t = \frac{- n + 4 m}{m + n}$ , and $- 1 + 3 t = \frac{2 n - 5 m}{m + n}$

Eliminate $t : \frac{- n + 4 m}{15} = \frac{- 4 m + 3 n}{3} \Rightarrow 24 m = 16 n \Rightarrow m : n = 2 : 3$ .

Let $A = (1, 2)$ , $B = (- 2, 3)$ , and $P = (x, y)$ be the point where $A B$ meets the line $3 x + 4 y = 7$ .

Suppose $P$ divides $A B$ in the ratio $k : 1$ .

Coordinates of $P$ using section formula: $x = \frac{k . (- 2) + 1.1}{k + 1} = \frac{- 2 k + 1}{k + 1}$ , and $y = \frac{k .3 + 1.2}{k + 1} = \frac{3 k + 2}{k + 1}$

Substituting into the line equation $3 x + 4 y = 7 : 3. \frac{- 2 k + 1}{k + 1} + 4. \frac{3 k + 2}{k + 1} = 7$

$\Rightarrow \frac{- 6 k + 3 + 12 k + 8}{k + 1} = 7 \Rightarrow \frac{6 k + 11}{k + 1} = 7$

$\Rightarrow 6 k + 11 = 7. (k + 1) \Rightarrow 6 k + 11 = 7 k + 7 \Rightarrow k = 4$ .

Let $A = (3, - 1)$ , $B = (8, 9)$ , and $P = (x, y)$ be the point where the line $y - x + 2 = 0$ meets $A B$ . Suppose $P$ divides $A B$ in the ratio $k : 1$ .

Coordinates of $P$ using section formula: $x = \frac{k .8 + 1.3}{k + 1} = \frac{8 k + 3}{k + 1}$ , and $y = \frac{k .9 + 1. (- 1)}{k + 1} = \frac{9 k - 1}{k + 1}$

Substituting into the line equation $y - x + 2 = 0 : \frac{9 k - 1}{k + 1} - \frac{8 k + 3}{k + 1} + 2 = 0$

Combining terms: $\frac{9 k - 1 - 8 k - 3}{k + 1} + 2 = 0 \Rightarrow \frac{k - 4}{k + 1} + 2 = 0$

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

Let $A = (5, - 4)$ , $B = (3, - 2)$ , and $P$ divides $A B$ in the ratio $4 : 3$ .

Coordinates of $P$ using section formula: $x = \frac{3.5 + 4.3}{4 + 3} = \frac{15 + 12}{7} = \frac{27}{7}$ , and $y = \frac{3. - 4 + 4. - 2}{4 + 3} = - \frac{20}{7}$ .

Distance of $P$ from origin: $O P = \sqrt{(27 / 7)^{2} + (- 20 / 7)^{2}} = \sqrt{729 / 49 + 400 / 49} = \sqrt{1129 / 49} = \sqrt{1129} / 7$ .

Let the vertices be $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ , $C (x_{3}, y_{3})$ .

Midpoints of sides: $M_{A B} = (1, 1)$ , $M_{B C} = (2, 3)$ , $M_{C A} = (4, 1)$

Midpoint formula: $M_{A B} = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) = (1, 1) \Rightarrow x_{1} + x_{2} = 2, y_{1} + y_{2} = 2$

$M_{B C} = (\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2}) = (2, 3) \Rightarrow x_{2} + x_{3} = 4, y_{2} + y_{3} = 6$

$M_{C A} = (\frac{x_{3} + x_{1}}{2}, \frac{y_{3} + y_{1}}{2}) = (4, 1) \Rightarrow x_{3} + x_{1} = 8, y_{3} + y_{1} = 2$

Solving for $x$ -coordinates: $x_{1} + x_{2} = 2, x_{2} + x_{3} = 4, x_{3} + x_{1} = 8$

Adding first two: $x_{1} + 2 x_{2} + x_{3} = 6$

Subtracting third: $(x_{1} + 2 x_{2} + x_{3}) - (x_{3} + x_{1}) = 6 - 8 \Rightarrow 2 x_{2} - x_{3} = - 2 \Rightarrow x_{3} = 2 x_{2} + 2$

From $x_{2} + x_{3} = 4 \Rightarrow x_{2} + 2 x_{2} + 2 = 4 \Rightarrow 3 x_{2} = 2 \Rightarrow x_{2} = 2 / 3$ Then $x_{3} = 2. (2 / 3) + 2 = 10 / 3$ , $x_{1} = 2 - x_{2} = 4 / 3$

Solving for $y$ -coordinates: $y_{1} + y_{2} = 2, y_{2} + y_{3} = 6, y_{3} + y_{1} = 2$

Adding first two: $y_{1} + 2 y_{2} + y_{3} = 8$

Subtracting third: $(y_{1} + 2 y_{2} + y_{3}) - (y_{3} + y_{1}) = 8 - 2 \Rightarrow 2 y_{2} = 6 \Rightarrow y_{2} = 3$

Then $y_{1} = 2 - 3 = - 1, y_{3} = 2 - y_{1} = 3$ .

The solutions are given below:

Let $A = (2, 4)$ , $B = (6, 4)$ , $C = (2, 0)$ .
Centroid $G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$
$G = (\frac{2 + 6 + 2}{3}, \frac{4 + 4 + 0}{3}) = (10 / 3, 8 / 3)$
Side lengths: $A B = \sqrt{(6 - 2)^{2} + (4 - 4)^{2}} = \sqrt{16} = 4$ ,
$B C = \sqrt{(6 - 2)^{2} + (4 - 0)^{2}} = \sqrt{16 + 16} = \sqrt{32} = 4 \sqrt{2}$ and
$C A = \sqrt{(2 - 2)^{2} + (0 - 4)^{2}} = \sqrt{16} = 4$ .
Incenter formula: $I = (\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c})$
Here, $a = B C = 4 \sqrt{2}$ , $b = C A = 4$ , $c = A B = 4$
$x$ -coordinate: $I_{x} = \frac{a .2 + b .6 + c .2}{a + b + c} = \frac{4 \sqrt{2} .2 + 4.6 + 4.2}{4 \sqrt{2} + 4 + 4}$
$I_{x} = \frac{8 \sqrt{2} + 24}{4 \sqrt{2} + 8}$
$y$ -coordinate: $I_{y} = \frac{a .4 + b .4 + c .0}{a + b + c} = (16 \sqrt{2} + 16) / (4 \sqrt{2} + 8)$ .
Let $A = (1, 2)$ , $B = (2, 3)$ , $C = (3, 4)$ .
Centroid $G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$
$G = (\frac{1 + 2 + 3}{3}, \frac{2 + 3 + 4}{3}) = (6 / 3, 9 / 3) = (2, 3)$
Side lengths: $A B = \sqrt{(2 - 1)^{2} + (3 - 2)^{2}} = \sqrt{1 + 1} = \sqrt{2}$
$B C = \sqrt{(3 - 2)^{2} + (4 - 3)^{2}} = \sqrt{1 + 1} = \sqrt{2}$
$C A = \sqrt{(3 - 1)^{2} + (4 - 2)^{2}} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}$
Incenter formula: $I = (\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c})$
Let $a = B C = \sqrt{2}$ , $b = C A = 2 \sqrt{2}$ , $c = A B = \sqrt{2}$
$x$ -coordinate: $I_{x} = \frac{a .1 + b .2 + c .3}{a + b + c} = \frac{\sqrt{2} + 2 \sqrt{2} .2 + \sqrt{2} .3}{\sqrt{2} + 2 \sqrt{2} + \sqrt{2}}$
$I_{x} = \frac{\sqrt{2} + 4 \sqrt{2} + 3 \sqrt{2}}{(4 \sqrt{2}} = 2$
$y$ -coordinate: $I_{y} = \frac{a .2 + 2.3 + a .4}{a + b + c} = \frac{2 \sqrt{2} + 6 \sqrt{2} + 4 \sqrt{2}}{4 \sqrt{2}} = 3$

Let $A = (- 1, 4)$ , $B = (5, 2)$ , $C = (x, y)$ , and centroid $G = (0, - 3)$ .

Centroid formula: $G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{x})$

$x$ -coordinate: $0 = \frac{- 1 + 5 + x}{3} \Rightarrow 4 + x = 0 \Rightarrow x = - 4$

$y$ -coordinate: $- 3 = \frac{4 + 2 + y}{3} \Rightarrow 6 + y = - 9 \Rightarrow y = - 15$ .

Let $A = (1, 4)$ , $B = (4, 8)$ , and $P$ divides $A B$ externally in the ratio $m : 1$ .

Let $A P = 10$ and $A B = \sqrt{(4 - 1)^{2} + (8 - 4)^{2}} = \sqrt{9 + 16} = 5$

Since $A P = A B + B P$ , $P$ divides $A B$ externally i.e. ratio $m : 1 = 2 : 1$

Coordinates of $P$ using external division formula: $x = \frac{m .4 - 1}{m - 1} = \frac{2.4 - 1}{2 - 1} = 7$ , and $y = \frac{m .8 - 4}{m - 1} = \frac{2.8 - 4}{2 - 1} = 12$ .

Let $A = (3, 4), B = (- 4, 3), C = (8, 6)$ .

Area formula: $Δ = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | 3. (3 - 6) + (- 4) . (6 - 4) + 8. (4 - 3) | = 9 / 2$ .

Let the vertices be $A (- 3, 2), B (7, - 6), C (- 5, - 4), D (5, 4)$ , taken in order.

Area formula for quadrilateral: $Δ = \frac{1}{2} | x_{1} y_{2} + x_{2} y_{3} + x_{3} y_{4} + x_{4} y_{1} - (y_{1} x_{2} + y_{2} x_{3} + y_{3} x_{4} + y_{4} x_{1}) |$

Substituting values: $S_{1} = (- 3) . (- 6) + 7. (- 4) + (- 5) .4 + 5.2 = 18 - 28 - 20 + 10 = - 20$

$S_{2} = 2.7 + (- 6) . (- 5) + (- 4) .5 + 4. (- 3) = 14 + 30 - 20 - 12 = 12$

$Δ = (1 / 2) | - 20 - 12 | = (1 / 2) .32 = 16$ .

Let A = (6, 3), B = (-3, 5), C = (4, -2), P = (x, y).

Area of a triangle with vertices $(X_{1}, Y_{1}), (X_{2}, Y_{2}), (X_{3}, Y_{3})$ : $Δ = \frac{1}{2} | X_{1} . (Y_{2} - Y_{3}) + X_{2} . (Y_{3} - Y_{1}) + X_{3} . (Y_{1} - Y_{2}) |$

$Δ A B C = \frac{1}{2} | 6. (5 - (- 2)) + (- 3) . (- 2 - 3) + 4. (3 - 5) | = \frac{49}{2}$

$Δ P B C = \frac{1}{2} | x . (5 - (- 2)) + (- 3) . (- 2 - y) + 4. (y - 5) | = \frac{7}{2} | x + y - 2 |$

Now, ratio: $\frac{Δ P B C}{Δ A B C} = \frac{| x + y - 2 |}{7}$ .

Let $A = (3, 3), B = (h, 0), C = (0, k)$ .

Points are collinear if the area of $△ A B C$ is $0 : Δ A B C = \frac{1}{2} | x_{1} . (y_{2} - y_{3}) + x_{2} . (y_{3} - y_{1}) + x_{3} . (y_{1} - y_{2}) |$

Substituting values: $Δ A B C = \frac{1}{2} | 3. (0 - k) + h . (k - 3) + 0. (3 - 0) | = \frac{1}{2} | h . k - 3. h - 3. k |$

For collinearity, $Δ A B C = 0 \Rightarrow h . k - 3. h - 3. k = 0$

Divide by $h . k$ ( $h! = 0, k! = 0$ ): $1 - \frac{3}{k} - \frac{3}{h} = 0 \Rightarrow \frac{1}{h} + \frac{1}{k} = \frac{1}{3}$ .

Let $A = (x_{1}, y_{1})$ , $B = (x_{2}, y_{2})$ , $C = (x_{3}, y_{3})$ . Let $(x, y)$ be a point on the internal bisector of $∠ A$ . Let $A B = c$ , $A C = b$ .

The area of a triangle can be expressed using the determinant: $Δ (X, Y, Z) = | \begin{matrix} X_{x} & X_{y} & 1 \\ Y_{x} & Y_{y} & 1 \\ Z_{x} & Z_{y} & 1 \end{matrix} |$

Let $Δ P A B = | \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{matrix} |$ and let $Δ P A C = | \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} |$

By the angle bisector theorem, a point on the bisector divides the opposite side in the ratio of adjacent sides: $\frac{B P}{P C} = \frac{A B}{A C} = \frac{c}{b}$

The signed areas satisfy the same ratio: $\frac{Δ P A B}{Δ P A C} = - \frac{P C}{P B} = - \frac{b}{c}$

Cross multiplying: $b . Δ P A B + c . Δ P A C = 0$ .

Let the points be: $P (a) = (\frac{a^{3}}{a - 1}, \frac{a^{2} - 3}{a - 1}), Q (b) = (\frac{b^{3}}{b - 1}, \frac{b^{2} - 3}{b - 1}), R (c) = (\frac{c^{3}}{c - 1}, \frac{c^{2} - 3}{c - 1})$

Three points are collinear if the determinant vanishes: $| \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} | = 0$

Substitute $x_{i}$ and $y_{i}$ : $| \begin{matrix} \frac{a^{3}}{a - 1} & \frac{a^{2} - 3}{a - 1} & 1 \\ \frac{b^{3}}{b - 1} & \frac{b^{2} - 3}{b - 1} & 1 \\ \frac{c^{3}}{c - 1} & \frac{c^{2} - 3}{c - 1} & 1 \end{matrix} | = 0$

Multiplying each row by its denominator to simplify: $| \begin{matrix} a^{3} & a^{2} - 3 & a - 1 \\ b^{3} & b^{2} - 3 & b - 1 \\ c^{3} & c^{2} - 3 & c - 1 \end{matrix} |$

Expanding the determinant and simplifying (using factorization of cubic polynomials) gives: $a b c - (a b + b c + c a) + 3 (a + b + c) = 0$ .

Let the vertices of the triangle be $A = (4, - 8)$ , $B = (- 9, 7)$ , $C = (x, y)$ . The centroid $G = (1, 4)$ .

Coordinates of the centroid: $x_{G} = \frac{x_{1} + x_{2} + x_{3}}{3}, y_{G} = \frac{y_{1} + y_{2} + y_{3}}{3}$

Substitute values: $1 = \frac{4 - 9 + x}{3} \Rightarrow 1 = \frac{- 5 + x}{3} \Rightarrow x = 8 \Rightarrow 4 = \frac{- 8 + 7 + y}{3} \Rightarrow 4 = \frac{- 1 + y}{3} \Rightarrow y = 13$

So $C = (8, 13)$ .

Area of $△ A B C : Δ = \frac{1}{2} | x_{1} . (y_{2} - y_{3}) + x_{2} . (y_{3} - y_{1}) + x_{3} . (y_{1} - y_{2}) |$

Substituting values: $Δ = \frac{1}{2} | 4. (7 - 13) + (- 9) . (13 + 8) + 8. (- 8 - 7) | = 166.5$ .

Assume an equilateral triangle has vertices $A = (x_{1}, y_{1})$ , $B = (x_{2}, y_{2})$ , $C = (x_{3}, y_{3})$ with all coordinates rational.

Distance formula: $A B^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ ,

$B C^{2} = (x_{3} - x_{2})^{2} + (y_{3} - y_{2})^{2}$ , and

$C A^{2} = (x_{1} - x_{3})^{2} + (y_{1} - y_{3})^{2}$ .

For an equilateral triangle: $A B^{2} = B C^{2} = C A^{2} = d^{2}$

Consider the line $A B$ >. The perpendicular from $C$ to $A B$ must satisfy the formula for height of an equilateral triangle: $h = \frac{\sqrt{3}}{2} d$

Coordinates of $C$ satisfy the perpendicular distance formula from line $A B$ : $\frac{| (y_{2} - y_{1}) x_{C} - (x_{2} - x_{1}) y_{C} + (x_{2} y_{1} - x_{1} y_{2})}{\sqrt{(y_{2} - y_{1})^{2} + (x_{2} - x_{1})^{2}}} = h = \frac{\sqrt{3}}{2} d$

All numbers on the left are rational, but the right-hand side involves $\sqrt{3}$ , which is irrational. This is a contradiction.

Hence, the coordinates of the vertices of an equilateral triangle cannot all be rational.

Let $A = (- 1, 5)$ , $B = (3, 1)$ , $C = (5, 7)$ .

Midpoints: $D = (\frac{3 + 5}{2}, \frac{1 + 7}{2}) = (4, 4)$ , $E = (\frac{5 - 1}{2}, \frac{7 + 5}{2}) = (2, 6)$ , and $F = (\frac{- 1 + 3}{2}, \frac{5 + 1}{2}) = (1, 3)$

$Δ A B C = \frac{1}{2} | (- 1) . (1 - 7) + 3. (7 - 5) + 5. (5 - 1) | = 16$

$Δ D E F = \frac{1}{2} | 4. (6 - 3) + 2. (3 - 4) + 1. (4 - 6) | = 4$

Hence, $Δ A B C = 4 Δ D E F$ .

Let $A = (3, 0)$ , $B = (0, 6)$ , $C = (6, 9)$ .

Points dividing sides: $D =$ divides $A B$ in $1 : 2$ $= (\frac{2.3 + 1.0}{3}, \frac{2.0 + 1.6}{3}) = (2, 2)$ > and $E =$ divides $A C$ in $1 : 2 = (\frac{2.3 + 1.6}{3}, \frac{2.0 + 1.9}{3}) = (4, 3)$

$Δ = \frac{1}{2} | x_{1} . (y_{2} - y_{3}) + x_{2} . (y_{3} - y_{1}) + x_{3} . (y_{1} - y_{2}) |$

$Δ A B C = \frac{1}{2} | 3. (6 - 9) + 0. (9 - 0) + 6. (0 - 6) | = 22.5$

$Δ A D E = \frac{1}{2} | 3. (2 - 3) + 2. (3 - 0) + 4. (0 - 2) | = 2.5$

$\frac{Δ A B C}{Δ A D E} = 9$ .

Let the vertices be: $A = (t, t - 2)$ , $B = (t + 3, t)$ , $C = (t + 2, t + 2)$ .

$Δ = \frac{1}{2} | x_{1} . (y_{2} - y_{3}) + x_{2} . (y_{3} - y_{1}) + x_{3} . (y_{1} - y_{2}) |$

Substitute coordinates: $Δ = \frac{1}{2} | t . (t - (t + 2)) + (t + 3) . ((t + 2) - (t - 2)) + (t + 2) . ((t - 2) - t) |$

$t . (t - t - 2) = t . (- 2) = - 2 t$ , $(t + 3) . (t + 2 - t + 2) = (t + 3) .4 = 4 t + 12$ , and $(t + 2) . (t - 2 - t) = (t + 2) . (- 2) = - 2 t - 4$ .

Adding terms: $- 2 t + 4 t + 12 - 2 t - 4 = 8$

Divide by $2$ : $Δ = \frac{1}{2} .8 = 4$ , which is independent of $t$ .

Let the vertices be $A = (x, y)$ , $B = (1, 2)$ , $C = (2, 1)$ .

Area formula: $Δ = \frac{1}{2} | x . (y_{2} - y_{3}) + x_{2} . (y_{3} - y) + x_{3} . (y - y_{2}) |$

Substitute coordinates: $6 = \frac{1}{2} | x . (2 - 1) + 1. (1 - y) + 2. (y - 2) |$

Simplify: $6 = \frac{1}{2} | x .1 + 1. (1 - y) + 2. (y - 2) | \Rightarrow 6 = \frac{1}{2} | x + y - 3 |$

Multiply both sides by $2 : 12 = | x + y - 3 |$

Solving for absolute value: $x + y - 3 = 12 \Rightarrow x + y = 15$ or $x + y - 3 = - 12 \Rightarrow x + y = - 9$ .

Vertices: $A = (1, 1)$ , $B = (7, 3)$ , $C = (12, 2)$ , $D = (7, 21)$ .

Area: $Δ = \frac{1}{2} | 1. (3 - 21) + 7. (2 - 1) + 12. (21 - 3) + 7. (1 - 2) | = 99$ .

Vertices: $A = (4, 3)$ , $B = (- 5, 6)$ , $C = (0, - 7)$ , $D = (3, - 6)$ , $E = (- 7, - 2)$ .

Area: $Δ = \frac{1}{2} | 4.6 + (- 5) . (- 7) + 0. (- 6) + 3. (- 2) + (- 7) .3 - (3. (- 5) + 6.0 + (- 7) .3 + (- 6) . (- 7) + (- 2) .4) | = 17$ .

Vertices: $A = (5, 0)$ , $B = (4, 2)$ , $C = (1, 3)$ , $D = (- 2, 2)$ , $E = (- 3, - 1)$ , $F = (0, - 4)$ .

Area: $Δ = \frac{1}{2} | 5.2 + 4.3 + 1.2 + (- 2) . (- 1) + (- 3) . (- 4) + 0.0 - (0.4 + 2.1 + 3. (- 2) + 2. (- 3) + (- 1) .0 + (- 4) .5) | = 34$ .

Vertices: $A = ((a + 1) (a + 2), a + 2)$ , $B = ((a + 2) (a + 3), a + 3)$ , $C = ((a + 3) (a + 4), a + 4)$ .

Area: $Δ = \frac{1}{2} | (a + 1) (a + 2) . ((a + 3) - (a + 4)) + (a + 2) (a + 3) . ((a + 4) - (a + 2))$ $+ (a + 3) (a + 4) . ((a + 2) - (a + 3)) | = 0$ .

Thus, the points are collinear.

A divides $P (- 5, 1)$ and $Q (3, 5)$ in ratio $k : 1$ : $A = (\frac{3 k - 5}{k + 1}, \frac{5 k + 1}{k + 1})$

Area of $△ A B C = 2$ : $2 = 1 / 2 | (\frac{3 k - 5}{k + 1}) . (5 - (- 2)) + 1. (- 2 - \frac{5 k + 1}{k + 1}) + 7. (\frac{5 k + 1}{k + 1} - 5) |$

$\Rightarrow k = 7$ or $k = \frac{31}{9}$ .

Vertices: $A = (6, 3), B = (- 3, 5), C = (4, - 2), D = (x, 3 x)$ .

Area formula: $Δ A B C = \frac{1}{2} | 6. (5 - (- 2)) + (- 3) . (- 2 - 3) + 4. (3 - 5) | = 24.5$

Area $Δ B C D = (1 / 2) | (- 3) . (- 2 - 3 x) + 4. (3 x - 5) + x . (5 - (- 2)) | = | 14 x - 7 |$

Given $Δ A B C = 2 Δ B C D \Rightarrow 24.5 = 2. | 14 x - 7 | \Rightarrow | 14 x - 7 | = 12.25$

Solving: $14 x - 7 = 12.25 \Rightarrow 14 x = 19.25 \Rightarrow x = 19.25 / 14 = \frac{77}{56}$

$14 x - 7 = - 12.25 \Rightarrow 14 x = - 5.25 \Rightarrow x = - \frac{3}{8}$ .

Vertices: $A = (1, 2), B = (- 5, 6), C = (7, - 4), D = (h, - 2)$ .

$Δ = 0 \Rightarrow 1. (6 - (- 4)) + (- 5) . (- 4 - (- 2)) + 7. (- 2 - 2) + h . (2 - 6) = 0$

$\Rightarrow h = 3$ .

Vertices: $A = (3, 4), B = (- 4, 3), C = (8, 6)$ .

Area: $Δ = \frac{1}{2} | 3. (3 - 6) + (- 4) . (6 - 4) + 8. (4 - 3) | = \frac{9}{2}$ .

Length of perpendicular from $A$ on $B C$ : $h = \frac{2 Δ}{B C}$

$B C = \sqrt{(8 + 4)^{2} + (6 - 3)^{2}} = \sqrt{153}$

$h = (2. (9 / 2)) / \sqrt{153} = 9 / \sqrt{153}$ .

Centroid $G = (2 / 3, 2)$ , vertices $A = (2, 3), B = (- 1, 2), C = (x, y)$ .

Centroid formula: $G = (\frac{x_{A} + x_{B} + x_{C}}{3}, \frac{y_{A} + y_{B} + y_{C}}{3}) \Rightarrow (\frac{2}{3}, 2) = (\frac{2 - 1 + x}{3}, \frac{3 + 2 + y}{3})$

Solve for $C$ : $\frac{2 - 1 + x}{3} = \frac{2}{3} \Rightarrow x = 1$ and $\frac{3 + 2 = y}{3} = 2 \Rightarrow y = - 1$

Area: $Δ = \frac{1}{2} | 2. (2 - (- 1)) + (- 1) . (- 1 - 3) + 1. (3 - 2) | = \frac{11}{2}$ .

Vertices: $A = (3, 1), B = (1, - 3), C = (x, y)$ , centroid $G = (\frac{3 + 1 + x}{3}, \frac{1 - 3 = y}{3})$ M lies on $x$ -axis $\Rightarrow y$ -coordinate $= 0$

$\frac{1 - 3 + y}{3} = 0 \Rightarrow y = 2$

Area formula: $Δ = 3 = \frac{1}{2} | 3. (- 3 - 2) + 1. (2 - 1) + x . (1 - (- 3)) | \Rightarrow x = 5$ , or $- 14 + 4 x = - 6 \Rightarrow x = 2$

So $C = (5, 2)$ or $C = (2, 2)$ .

Given: $A = (3, 4), B = (5, - 2), P = (x, y)$ such that $P A = P B$

Perpendicular bisector condition: $(x - 3)^{2} + (y - 4)^{2} = (x - 5)^{2} + (y + 2)^{2}$

Simplifying: $x^{2} - 6 x + 9 + y^{2} - 8 y + 16 = x^{2} - 10 x + 25 + y^{2} + 4 y + 4$

$4 x - 12 y = - 12 => x - 3 y = - 3$

Area formula: $Δ P A B = 10 = \frac{1}{2} | 3 (- 2 - y) + 5 (y - 4) + x (4 - (- 2)) |$

Simplifying: $20 = | - 6 - 3 y + 5 y - 20 + 6 x | = | 6 x + 2 y - 26 |$

Solve system: $x - 3 y = - 3$ and $6 x + 2 y - 26 = \pm 20$

Case I: $6 x + 2 y - 26 = 20 \Rightarrow 6 x + 2 y = 46 \Rightarrow 3 x + y = 23$

Solving with $x - 3 y = - 3 \Rightarrow x = \frac{17}{2}, y = \frac{11}{3}$

Case II: $6 x + 2 y - 26 = - 20 \Rightarrow 6 x + 2 y = 6 \Rightarrow 3 x + y = 3$

Solving with $x - 3 y = - 3 \Rightarrow x = - \frac{3}{2}, y = - \frac{1}{2}$ .

Let the points be $A = (a, b + c), B = (b, c + a), C = (c, a + b)$ .

Area of triangle: $Δ = \frac{1}{2} | a ((c + a) - (a + b)) + b ((a + b) - (b + c)) + c ((b + c) - (c + a)) |$

Simplifying each term: $((c + a) - (a + b)) = c - b$ , $((a + b) - (b + c)) = a - c$ , $((b + c) - (c + a)) = b - a$

So $Δ = 1 / 2 | a (c - b) + b (a - c) + c (b - a) |$

Expanding: $a (c - b) + b (a - c) + c (b - a) = a c - a b + a b - b c + b c - a c = 0$ .

Hence, $Δ = 0$ i.e. points are collinear.

Let the points be $A = (x_{1}, y_{1}), B = (x_{2}, y_{2}), C = (x_{3}, y_{3})$ .

If points are collinear, $Δ = 0 : Δ = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) | = 0$

Divide both sides by $x_{1} x_{2} x_{3} \neq 0$ : $0 = \frac{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})}{x_{1} x_{2} x_{3}}$

Splitting terms: $0 = \frac{y_{2} - y_{3}}{x_{2} x_{3}} + \frac{y_{3} - y_{1}}{x_{3} x_{1}} + \frac{y_{1} - y_{2}}{x_{1} x - 2}$

Hence proved.

Points: $A = (a, b), B = (a_{1}, b_{1}), C = (a - a_{1}, b - b_{1})$ .

Collinear $\Rightarrow Δ = 0 \Rightarrow 0 = \frac{1}{2} | a (b_{1} - (b - b_{1})) + a_{1} ((b - b_{1}) - b) + (a - a_{1}) ((b - b_{1}) - b_{1}) |$

Simplifying: $a_{1} b = a b_{1} \Rightarrow \frac{a}{a_{1}} = \frac{b}{b_{1}}$ .

Points: $A = (a, 0), B = (0, b), C = (1, 1)$ .

Collinear $\Rightarrow Δ = 0 \Rightarrow 0 = \frac{1}{2} | a (b - 1) + 0 (1 - 0) + 1 (0 - b) |$

Simplify: $0 = a (b - 1) - b \Rightarrow a b - a - b = 0 \Rightarrow \frac{1}{a} + \frac{1}{b} = 1$

Hence, points are collinear if $\frac{1}{a} + \frac{1}{b} = 1$ .

Let the points be $A = (- 4, - 1), B = (- 2, - 4), C = (4, 0), D = (2, 3)$ .

$A B = \sqrt{(- 2 + 4)^{2} + (- 4 + 1)^{2}} = \sqrt{13}$

$B C = \sqrt{(4 + 2)^{2} + (0 + 4)^{2}} = \sqrt{52}$

$C D = \sqrt{(2 - 4)^{2} + (3 - 0)^{2}} = \sqrt{13)}$

$D A = \sqrt{(- 4 - 2)^{2} + (- 1 - 3)^{2}} = \sqrt{52}$

Opposite sides equal: $A B = C D, B C = D A$ .

Diagonals: $A C = \sqrt{(4 + 4)^{2} + (0 + 1)^{2}} = \sqrt{65}$

$B D = \sqrt{(2 + 2)^{2} + (3 + 4)^{2}} = \sqrt{65}$

Equal diagonals imply that all angles are right angles.

Let the three consecutive vertices be $A = (x_{1}, y_{1}), B = (x_{2}, y_{2}), C = (x_{3}, y_{3})$ , and let $D = (x_{4}, y_{4})$ be the fourth vertex.

In a parallelogram, the diagonals bisect each other $\Rightarrow \frac{x_{1} + x_{3}}{2} = \frac{x_{2} + x_{4}}{2}$ and $\frac{y_{1} + y_{3}}{2} = \frac{y_{2} + y_{4}}{2}$

Solving for $D : x_{4} = x_{1} + x_{3} - x_{2}$ and $y_{4} = y_{1} + y_{3} - y_{2}$

Hence, the fourth vertex is $D = (x_{1} + x_{3} - x_{2}, y_{1} + y_{3} - y_{2})$ .

Let $B = (0, 0), C = (c, 0)$ and $D$ be the midpoint of $B C \Rightarrow D = (\frac{c}{2}, 0)$ . Let $A = (x, y)$ .

$A B^{2} = (x - 0)^{2} + (y - 0)^{2} = x^{2} + y^{2}$

$A C^{2} = (x - c)^{2} + (y - 0)^{2} = (x - c)^{2} + y^{2}$

$B D^{2} = (c / 2 - 0)^{2} + (0 - 0)^{2} = \frac{c^{2}}{4}$

$A D^{2} = {(x - \frac{c}{2})}^{2} + (y - 0)^{2} = {(x - \frac{c}{2})}^{2} + y^{2}$

Left-hand side: $A B^{2} + A C^{2} = x^{2} + y^{2} + (x - c)^{2} + y^{2} = 2 x^{2} - 2 c x + c^{2} + 2 y^{2}$

Right-hand side: $2 (A D^{2} + B D^{2}) = 2 ({(x - \frac{c}{2})}^{2} + y^{2} + \frac{c^{2}}{4}) = 2 (x^{2} - c x + \frac{c^{2}}{4} + y^{2} + \frac{c^{2}}{4}) = 2 x^{2} - 2 c x + c^{2} + 2 y^{2}$

Hence, $A B^{2} + A C^{2} = 2 (A D^{2} + B D^{2})$ .

Let $A = (x_{1}, y_{1}), B = (x_{2}, y_{2}), C = (x_{3}, y_{3})$ and $G$ be the centroid: $G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$

Let $O = (x_{0}, y_{0})$ .

Using distance formula: $O A^{2} + O B^{2} + O C^{2} = (x_{0} - x_{1})^{2} + (y_{0} - y_{1})^{2} + (x_{0} - x_{2})^{2} + (y_{0} - y_{2})^{2} + (x_{0} - x_{3})^{2} + (y_{0} - y_{3})^{2}$

Grouping terms: $= 3 x_{0}^{2} - 2 x_{0} (x_{1} + x_{2} + x_{3}) + (x_{1}^{2} + x_{2}^{2} + x_{3}^{2}) + 3 y_{0}^{2} - 2 y_{0} (y_{1} + y_{2} + y_{3}) + (y_{1}^{2} + y_{2}^{2} + y_{3}^{2})$

Since $G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ , rewriting: $O A^{2} + O B^{2} + O C^{2} = G A^{2} + G B^{2} + G C^{2} + 3 G O^{2}$ .

Let $A = (x_{1}, y_{1}), B = (x_{2}, y_{2}), C = (x_{3}, y_{3})$ , and let $D, E, F$ be the midpoints of $B C, C A, A B$ respectively:

$D = (\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2}), E = (\frac{x_{3} + x_{1}}{2}, \frac{y_{3} + y_{1}}{2}), F = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

$Δ_{A B C} = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$Δ_{D E F} = \frac{1}{2} | D_{x} (E_{y} - F_{y}) + E_{x} (F_{y} - D_{y}) + F_{x} (D_{y} - E_{y}) |$

Substituting midpoints: $Δ_{D E F} = \frac{1}{2} . \frac{1}{2^{2}} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) | = 1 / 4 Δ_{A B C}$

Let $A = (x_{1}, y_{1}), B = (x_{2}, y_{2}), C = (x_{3}, y_{3})$ be the vertices of a triangle. Let $D$ and $E$ be the midpoints of $A B$ and $A C$ :

$D = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}), E = (\frac{x_{1} + x_{3}}{2}, \frac{y_{1} + y_{3}}{2})$ .

Length of $D E : D E^{2} = {(\frac{x_{1} + x_{3}}{2} - \frac{x_{1} + x_{2}}{2})}^{2} + {(\frac{y_{1} + y_{3}}{2} - \frac{y_{1} + y_{2}}{2})}^{2}$

$= {(\frac{x_{3} - x_{2}}{2})}^{2} + {(\frac{y_{3} - y_{2}}{2})}^{2}$

But $B C^{2} = (x_{3} - x_{2})^{2} + (y_{3} - y_{2})^{2} \Rightarrow D E = \frac{1}{2} B C$ .

Let $A = (x_{1}, y_{1}), B = (x_{2}, y_{2}), C = (x_{3}, y_{3})$ .

Let $P$ divide $B C$ in ratio $k : 1 \Rightarrow P = (\frac{k x_{3} + x_{2}}{k + 1}, \frac{k y_{3} + y_{2}}{k + 1})$

Let $Q$ divide $C A$ in ratio $k : 1 \Rightarrow Q = (\frac{k x_{1} + x_{3}}{k + 1}, \frac{k y_{1} + y_{3}}{k + 1})$

Let $R$ divide $A B$ in ratio $k : 1 \Rightarrow R = (\frac{k x_{2} + x_{1}}{k + 1}, \frac{k y_{2} + y_{1}}{k + 1})$

Centroid of $A B C : G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$

Centroid of $P Q R : G^{'} = (\frac{\frac{k x_{3} + x_{2}}{k + 1} + \frac{k x_{1} + x_{3}}{k + 1} + \frac{k x_{2} + x_{1}}{k + 1}}{3}, \frac{\frac{k y_{3} + y_{2}}{k + 1} + \frac{k y_{1} + y_{3}}{k + 1} + \frac{k y_{2} + y_{1}}{k + 1}}{3})$

Simplifying $G^{'}$ gives $G^{'} = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}) = G$ .

Hence, the centroids coincide.

Let $A = (x_{1}, y_{1}), B = (x_{2}, y_{2}), C = (x_{3}, y_{3})$ . Let $D, E, F$ be the midpoints of $B C, C A, A B$ respectively.

$D = (\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2}), E = (\frac{x_{3} + x_{1}}{2}, \frac{y_{3} + y_{1}}{2})$ , and $F = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

Medians: $A D, B E, C F$

Using distance formula and simplifying: $A D^{2} = \frac{1}{4} [2 (A B^{2} + A C^{2}) - B C^{2}]$

$B E^{2} = \frac{1}{4} [2 (B C^{2} + A B^{2}) - C A^{2}]$

$C F^{2} = \frac{1}{4} [2 (C A^{2} + B C^{2}) - A B^{2}]$

$\Rightarrow A D^{2} + B E^{2} + C F^{2} = \frac{1}{4} [3 (A B^{2} + B C^{2} + C A^{2})]$

$\Rightarrow 4 (A D^{2} + B E^{2} + C F^{2}) = 3 (A B^{2} + B C^{2} + C A^{2})$ .

Let $A = (x_{1}, y_{1}), B = (x_{2}, y_{2}), C = (x_{3}, y_{3})$ . Centroid $G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ .

$G A^{2} + G B^{2} + G C^{2} =$
$\frac{1}{3} [(x_{1} - x_{2})^{2} + (x_{2} - x_{3})^{2} + (x_{3} - x_{1})^{2} + (y_{1} - y_{2})^{2} + (y_{2} - y_{3})^{2} + (y_{3} - y_{1})^{2}]$

But, $A B^{2} + B C^{2} + C A^{2} = (x_{1} - x_{2})^{2} + (x_{2} - x_{3})^{2} + (x_{3} - x_{1})^{2} + (y_{1} - y_{2})^{2} + (y_{2} - y_{3})^{2} + (y_{3} - y_{1})^{2}$

Hence, $A B^{2} + B C^{2} + C A^{2} = 3 (G A^{2} + G B^{2} + G C^{2})$ .

Let $A = (x_{1}, y_{1}), B = (x_{2}, y_{2}), C = (x_{3}, y_{3})$ . Centroid $G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ .

Area of $△ A B G : Δ A B G = \frac{1}{2} | x_{1} (y_{2} - y_{G}) + x_{2} (y_{G} - y_{1}) + x_{G} (y_{1} - y_{2}) |$

Substitute $x_{G}, y_{G}$ and simplify: $Δ A B G = \frac{1}{3} Δ A B C$

Similarly, $Δ B C G = \frac{1}{3} Δ A B C$ and $Δ C A G = \frac{1}{3} Δ A B C$

Hence, $Δ A B G = Δ B C G = Δ C A G$ .

Let the right angled triangle be right angled at $a$ . Take $A = (0, 0), B = (a, 0), C = (0, b)$ .

Midpoint of hypotenuse $B C : D = (\frac{a + 0}{2}, \frac{0 + b}{2}) = (\frac{a}{2}, \frac{b}{2})$

Distances: $D A^{2} = {(\frac{a}{2})}^{2} + {(\frac{b}{2})}^{2} = \frac{a^{2} + b^{2}}{4}$

$D B^{2} = {(\frac{a}{2} - a)}^{2} + {(\frac{b}{2} - 0)}^{2} = \frac{a^{2} + b^{2}}{4}$

$D C^{2} = {(\frac{a}{2} - 0)}^{2} + {(\frac{b}{2} - b)}^{2} = \frac{a^{2} + b^{2}}{4}$

Thus, $D A = D B = D C$ .

Let $A = (a_{1}, b_{1}), B = (a_{2}, b_{2}), C = (a_{3}, b_{3})$ .

$D$ divides $A B$ in ratio $λ : μ : D = (\frac{λ . a_{2} + μ . a_{1}}{λ + μ}, \frac{λ . b_{2} + μ . b_{1}}{λ + μ})$

$E$ divides $D C$ in ratio $m u : (λ + μ) : E = (\frac{μ . a_{3} + (λ + μ) . \frac{λ . a_{2} + μ . a_{1}}{λ + μ}}{λ + 2 μ}, \frac{μ . b_{3} + (λ + μ) \frac{λ . b_{2} + μ . b_{1}}{λ + μ}}{λ + 2 μ})$

Simplifying: $E = (\frac{λ . a_{2} + μ . a_{1} + μ . a_{3}}{λ + 2 μ}, \frac{λ . b_{2} + μ . b_{1} + μ . b_{3}}{λ + 2 μ})$ .

Given roots: $α + β = - \frac{2 h}{a}, α . β = \frac{b}{a}$ and $γ + δ = - \frac{2 h^{'}}{a}, γ . δ = \frac{b^{'}}{a}$

Ratios in which $C$ and $D$ divide $A B : \frac{A C}{C B} = \frac{γ - α}{β - γ}$ and $\frac{A D}{D B} = \frac{δ - α}{β - δ}$

Sum: $S = \frac{γ - α}{β - γ} + \frac{δ - α}{β - d e l t a}$

Simplify numerator: $S = \frac{(γ - α) (β - δ) + (δ - α) (β - γ)}{(β - γ) (β - δ)}$

Numerator expands to: $β (γ + δ) - α (γ + δ) - 2 γ . δ + 2 α . β$

Substitute root relations: $= β . (- \frac{2 h^{'}}{a}) - α (- \frac{2 h^{'}}{a}) - 2 \frac{b^{'}}{a} + 2 \frac{b}{a} = \frac{2 h^{'}}{a} (α - β) + 2 \frac{b - b^{'}}{a}$

Using condition $a . b^{'} + a^{'} . b = 2 h . h^{'}$ gives numerator $= 0$

Hence: $S = 0$

Thus, the sum of the ratios in which $C$ and $D$ divide $A B$ is zero.

Given $A = (1, - 2), B = (2, 5)$ and $O = (0, 0)$ .

$O C = 2 O A \Rightarrow C = (2.1, 2. (- 2)) = (2, - 4)$ and $O D = 2 O B \Rightarrow D = (2.2, 2.5) = (4, 10)$

Distance $C D : C D = \sqrt{(4 - 2)^{2} + (10 - (- 4))^{2}} = \sqrt{2^{2} + 14^{2}} = 10 \sqrt{2}$

Given $A = (2, 1), B = (3, - 1)$ and $C = (x, y)$ with $y = x + 9$ .

Centroid $G = (\frac{2 + 3 + x}{3}, \frac{1 - 1 + y}{3})$ lies on $y$ -axis $\Rightarrow x$ -coordinate of $G = 0 : \frac{5 + x}{3} = 0 => x = - 5$

Then $y = x + 9 = 4 \Rightarrow C = (- 5, 4)$ .

Centroid: $G = (\frac{2 + 3 - 5}{3}, \frac{1 - 1 + 4}{3}) = (0, \frac{4}{3})$

Let the vertices of the triangle be $A = (x_{1}, y_{1}), B = (x_{2}, y_{2}), C = (x_{3}, y_{3})$ .

Centroid: $(x, y) = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$

Circumcenter: $(α, β)$ . Orthocenter: $(p, q)$ .

In coordinate geometry, the orthocenter satisfies: $p = x_{1} + x_{2} + x_{3} - 2 α$ and x1 $q = y_{1} + y_{2} + y_{3} - 2 β$

Multiplying centroid coordinates by $3 : 3 x = x_{1} + x_{2} + x_{3}$ and $3 y = y_{1} + y_{2} + y_{3}$

Substitute: $3 x = 2 α + p$ and $3 y = 2 β + q$ . Hence proved.

Vertices: $A = (2, 3), B = (3, 4), C = (6, 8)$

Centroid $G : G = (\frac{2 + 3 + 6}{3}, \frac{3 + 4 + 8}{3}) = (\frac{11}{3}, 5)$

Circumcenter $(α, β)$ : Using the perpendicular bisector formula: the circumcenter is the intersection of the perpendicular bisectors of any two sides.

Equation of perpendicular bisector of $A B$ : Midpoint $M_{1} = (\frac{2 + 3}{2}, \frac{3 = 4}{2}) = (\frac{5}{2}, \frac{7}{2})$

If $C = (6, 8)$ , line $A C$ : slope not needed, solve using: $(x - α)^{2} + (y - β)^{2} = (x_{C} - α)^{2} + (y_{C} - β)^{2}$

$(2 - α)^{2} + (3 - β)^{2} = (3 - α)^{2} + (4 - β)^{2} \Rightarrow (2 - α)^{2} + (3 - β)^{2} = (6 - α)^{2} + (8 - β)^{2}$

Solve these two equations simultaneously: $α = \frac{9}{2}, β = \frac{3}{2}$

Orthocenter $(p, q)$ using formula $3 x = 2 α + p, 3 y = 2 β + q : p = 3. \frac{11}{3} - 2. \frac{9}{2} = 2$ and $q = 3.5 - 2. \frac{3}{2} = 12$ .

Let $A (α, 1 / α), B (β, 1 / β), C (γ, 1 / γ)$ be the vertices of a triangle.

$α$ and $β$ are roots of $x^{2} - 6 p_{1} x + 2 = 0$ , so $α + β = 6 p_{1}$ and $α β = 2$ .

$β$ and $γ$ are roots of $x^{2} - 6 p_{2} x + 3 = 0$ , so $β + γ = 6 p_{2}$ and $β γ = 3$ .

$γ$ and $α$ are roots of $x^{2} - 6 p_{3} x + 6 = 0$ , so $γ + α = 6 p_{3}$ and $γ α = 6$ .

Solving these gives $p_{1} = 1, p_{2} = 1, p_{3} = 2, α = 2, β = 1, γ = 3$ .

The centroid of the triangle is $(\frac{α + β + γ}{3}, \frac{\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}}{3}) = (\frac{2 + 1 + 3}{2}, \frac{\frac{1}{2} + 1 + \frac{1}{3}}{3}) = (2, \frac{11}{18})$ .

Let the vertices of the triangle be $A (\tan α, \cot α)$ $B (\tan β, \cot β)$ $C (\tan γ, \cot γ)$ , where $\tan α, \tan β, \tan γ$ are roots of $x^{3} - 3 a x^{2} + 3 b x - 1 = 0$ .

By Vieta's formulas, $\tan α + \tan β + \tan γ = 3 a$ ,

$\tan α . \tan β + \tan β . \tan γ + \tan γ . \tan α = 3 b$ , and $\tan α . \tan β . \tan γ = 1$ .

The centroid of the triangle is $(\frac{\tan α + \tan β + \tan γ}{3}, \frac{\cot α + \cot β + \cot γ}{3}) = (\frac{3 a}{3}, \frac{\cot α + \cot β + \cot γ}{3})$ .

Since $\cot θ = \frac{1}{\tan θ}$ , we have $\cot α + \cot β + \cot γ = \frac{1}{\tan α} + \frac{1}{\tan β} + \frac{1}{\tan γ} = \frac{\tan β . \tan γ + \tan γ . \tan α + \tan α . \tan β}{\tan α . \tan β . \tan γ} = 3 b$ .

Therefore, the centroid is $(a, b)$ .

Let the forces $F_{1} = 30$ N and $F_{2} = 40$ N act at points $A (- 3, - 1)$ and $B (4, 6)$ .

The resultant acts along the line joining $A$ and $B$ and its point of application $R (x, y)$ divides $A B$ in the inverse ratio of the forces: $A R : R B = F_{2} : F_{1} = 40 : 30 = 4 : 3$ .

Using coordinate geometry, if a point divides the segment joining $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ in the ratio $m : n$ , its coordinates are $(\frac{n x_{1} + m x_{2}}{m + n}, \frac{n y - 1 + m y_{2}}{m + n})$ .

Here $x_{1} = - 3, y_{1} = - 1, x_{2} = 4, y_{2} = 6, m = 4, n = 3$ . Then

$x_{R} = \frac{3. (- 3) + 4.4}{3 + 4} = 1$ , and $y_{R} = \frac{3. (- 1) + 4.6}{3 + 4} = 3$ .

Thus, the point of application of the resultant force is $R (1, 3)$ .

Let $A (- 1, 3), B (- 2, 4)$ , and the other vertices be $C (x_{1}, y_{1}), D (x_{2}, y_{2})$ with diagonals intersecting at $O$ on the positive $x$ -axis.

For a parallelogram, diagonals bisect each other: $O = (\frac{- 1 + x_{2}}{2}, \frac{3 + y_{2}}{2}) = (\frac{- 2 + x_{1}}{2}, \frac{4 + y_{1}}{2})$ .

Let $O = (h, 0)$ , then $y_{1} = - 4, y_{2} = - 3, x_{1} = 2 h + 2, x_{2} = 2 h + 1$ .

Area formula: $| (x_{1} + 1) (y_{2} - 3) - (x_{2} + 1) (y_{1} - 3) | = 24$ .

Substituting: $| (2 h + 3) (- 6) - (2 h + 2) (- 7) | = | 2 h - 4 | = 24 \Rightarrow h = 14$ .

Hence, $C (30, - 4)$ and $D (29, - 3)$ .

Let $A (1, 2), B (8, 4), C (4, 10)$ .

The point $P (x, y)$ such that triangles $P C B, P C A$ , and $P A B$ have equal area is the centroid of $△ A B C$ :

$P = (\frac{1 + 8 + 4}{3}, \frac{2 + 4 + 10}{3}) = (\frac{13}{3}, \frac{16}{3})$ .

Let $a, b, c$ be the $p$ th, $q$ th, $r$ th terms of an H.P., so $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.

Consider the points $X (b c, p), Y (c a, q), Z (a b, r)$ .

Three points are collinear if $b c (q - r) + c a (r - p) + a b (p - q) = 0$ .

Since $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P., this relation is satisfied.

Hence the points $(b c, p), (c a, q), (a b, r)$ are collinear.

Let $x_{1}, x_{2}, x_{3}$ be in A.P. and $y_{1}, y_{2}, y_{3}$ be in A.P.

Consider the points $P_{1} (x_{1}, y_{1}), P_{2} (x_{2}, y_{2}), P_{3} (x_{3}, y_{3})$ .

Three points are collinear if the determinant $(x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})) = 0$ .

Since $x_{1}, x_{2}, x_{3}$ are in A.P., $x_{2} - x_{1} = x_{3} - x_{2}$ .

Since $y_{1}, y_{2}, y_{3}$ are in A.P., $y_{2} - y_{1} = y_{3} - y_{2}$ .

Then $x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = x_{1} (- (y_{2} - y_{1})) + x_{2} ((y_{2} - y_{1}) + (y_{2} - y_{1})) + x_{3} (- (y_{2} - y_{1})) = 0$ .

Hence the points $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , $(x_{3}, y_{3})$ are collinear.

Consider the points $(a, a^{2}), (b, b^{2}), (c, c^{2})$ .

Three points are collinear if $(a (b^{2} - c^{2}) + b (c^{2} - a^{2}) + c (a^{2} - b^{2})) = 0$ .

Factor each difference of squares: $a (b - c) (b + c) + b (c - a) (c + a) + c (a - b) (a + b) = 0$ .

Simplifying: $(a - b) (b - c) (c - a)! = 0$ for distinct $a, b, c$ .

Hence, the points $(a, a^{2}), (b, b^{2}), (c, c^{2})$ are not collinear.

Let $A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})$ be the vertices of a triangle, and let $M$ be the midpoint of $B C$ , so the median through $A$ is the line joining $A$ and $M$ .

Let $(x, y)$ be a point on this median. Then $M = (\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2})$ and $(x, y)$ lies on the line joining $A$ and $M$ :

$\frac{y - y_{1}}{\frac{y_{2} + y_{3}}{2} - y_{1}} = \frac{x - x_{1}}{\frac{x - 2 + x_{3}}{2} - x_{1}}$ .

Consider the determinants: $| \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{matrix} |$ and $| \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} |$

The first determinant represents twice the signed area of triangle $A X B$ and the second determinant represents twice the signed area of triangle $A X C$ .

Since $(x, y)$ lies on the median $A M$ , triangles $A X B$ and $A X C$ have equal areas with opposite signs, so their determinants sum to zero:

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

Let $A (2, - 3), B (3, - 2), C (x, y)$ .

Area $= \frac{3}{2} \Rightarrow | 2 (- 2 - y) + 3 (y + 3) + x (- 3 + 2) | / 2 = \frac{3}{2} \Rightarrow | x - y - 5 | = 3 \Rightarrow x - y = 8$ or $x - y = 2$ .

Centroid $G = (\frac{2 + 3 + x}{3}, \frac{- 3 - 2 + y}{3})$ lies on $3 x - y - 8 = 0 \Rightarrow 3 (\frac{5 + x}{3}) - \frac{- 5 + y}{3} - 8 = 0 \Rightarrow 3 x - y = - 2$ .

Solve with area condition: $x - y = 8$ and $3 x - y = - 2 \Rightarrow x = - 5, y = - 13$ .

Let $A (- 2, 5), B (4, - 1), C (9, 1), D (3, 7)$ .

Midpoint of $A C = (\frac{- 2 + 9}{2}, \frac{5 + 1}{2}) = (\frac{7}{2}, 3)$ , midpoint of $B D = (\frac{4 + 3}{2}, \frac{- 1 + 7}{2}) = (\frac{7}{2}, 3)$ , so $A B C D$ is a parallelogram.

Area $= | (- 2. - 1 + 4.1 + 9.7 + 3.5) - (5.4 + - 1.9 + 1.3 + 7. - 2) | / 2 = 42$ .

$E$ divides $A C$ in ratio $2 : 1 \Rightarrow E = (\frac{16}{3}, \frac{7}{3})$ , midpoint of $B C$ is $F = (\frac{13}{2}, 0)$ .

$Δ_{D E F} = 0$ , so $D, E, F$ are collinear.

Let the points be $A (- 3, - 1), B (2, - 1), C (1, 1), D (- 2, 1)$ .

Compute slopes of opposite sides:

Slope of $A B = \frac{- 1 + 1}{2 + 3}) = 0$ ,

Slope of $C D = \frac{1 - 1}{1 + 2} = 0$ .

Slope of $B C = \frac{1 + 1}{1 - 2} = - 2$ ,

Slope of $A D = \frac{1 + 1}{- 2 + 3} = 2$ .

Since $A B ∥ C D$ and $B C ∦ A D$ , the quadrilateral is a trapezium.

Let the vertices of a triangle be $A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})$ with integer coordinates.

The squared distance between two points is $A B^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ , similarly for $B C^{2}$ and $C A^{2}$ .

If the triangle is equilateral, then $A B^{2} = B C^{2} = C A^{2} = k^{2}$ for some integer $k^{2}$ .

Consider the triangle modulo $y$ : the square of any integer is $0$ or $1$ modulo $4$ . The sum of two squares modulo $1$ can be $0, 1, 2$ , but never $3$ . In an equilateral triangle with integer coordinates, all three squared distances must be equal modulo $4$ .

It can be shown that no three distinct integer points satisfy $(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} = (x_{3} - x_{2})^{2} + (y_{3} - y_{2})^{2} = (x_{1} - x_{3})^{2} + (y_{1} - y_{3})^{2}$ modulo $4$ . Hence, a triangle with integral coordinates cannot be equilateral.

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Chapter 7. Answers of Coordinates