The joint equation is given by $(x-4y+2)(x-y-1)=0\Rightarrow x^2-5xy+4y^2+x+2y-2=0$.

Let $y=mx$ be the straight line passing through origin and making an angle $\alpha$ with $y+x=0$.

Then $\tan \alpha =\frac{|m+1|}{1+m}\Rightarrow m^2(1-{\tan }^2\alpha )+2m(1+{\tan }^2\alpha )+(1-{\tan }^2\alpha )=0$

$\Rightarrow m_1+m_2=-2\sec 2\alpha ,m_1{m}_2=1$.

Thus, $(y-m_{1}x)(y-m_{2}x)=x^2+2xy\sec 2\alpha +y^2=0$.

Since the pair of lines are perpendicular and form an isosceles triangle with $2x+3y=6$ the pair will make ${45}^{\circ }$ with it. Let $m$ be the slope of pair of straight lines. Then

$\tan {45}^{\circ }=1=\left|\frac{m-(-\frac{2}{3})}{1+\frac{2m}{3}}\right|\Rightarrow m=1/5,-5$. So the lines are $5y-x=0$ and $5x+y=0$ as they pass through origin.

Solving the three lines pairwaise we obtain vertices as $(0,0),(\frac{20}{13},\frac{6}{13}),(-\frac{6}{13},\frac{30}{13})$.

Thus, $\mathrm{\Delta }=\frac{936}{169}$.

The combined equation is given by $(2x-y-3)(3x-y+4)=6x^2-5xy+y^2-x-y-12=0$.

Lines parallel to $x-2y=5$ and $x=3y-4$ are given by $x-2y=c$ and $x-3y=k$.

Given that both pass through $(1,2)$, thus, $c=-3$ and $k=-5$. So the lines are $x-2y+3=0$ and $x-3y+5=0$.

Thus, combined equation is $(x-2y+3)(x-3y+5)=x^2-5xy+6x^2+8x-19y+15=0$.

The bisectors of angle between coordinates axes is given by $x-y=0$ and $x+y=0$.

The combined equation will be therefore $x^2-y^2=0$.

Comparing given equation with general equation in second degree we have $a=8,b=2,c=15,2h=8,2g=26,2f=13$

Now $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$. Hence, the given equation represents a pair of straight lines.

Comparing the given equation with the general equation of second degree we get

$a=6,h=k,b=12,g=11,f=\frac{31}{2},c=20$

Putting these values in $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$ gives us

$40k^2-682k+2907=0\Rightarrow k=\frac{682\pm 2}{80}=\frac{171}{20},\frac{17}{2}$.

Comparing the given equation with the general equation of second degree we get

$a=10,2h=-11,b=-6,g=-6,f=-\frac{1}{2},c=2$

Putting these values in $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$ gives us

$0=0$. Hence, the give equation represents pair of straight lines.

Comparing the given equation with the general equation of second degree we get

$a=2,2h=-15,b=-17,g=2,f=\frac{23}{2},c=-6$

Putting these values in $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$ gives us

$0=0$. Hence, the give equation represents pair of straight lines.

Comparing the given equation with the general equation of second degree we get

$a=m,2h=-5,b=-6,2g=14,2f=5,c=4$

Putting these values in $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$ gives us

$m=6$. We see that $a+b=0$, hence, the lines are perpendicular to each other.

Comparing the given equation with the general equation of second degree we get

$a=1,2h=m,b=-2,2g=0,2f=3,c=-1$

Putting these values in $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$ gives us $m=\pm 1$.

Comparing the given equation with the general equation of second degree we get

$a=12,2h=-10,b=2,2g=11,2f=-5,c=m$

Putting these values in $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$ gives us $m=2$.

Comparing the given equation with the general equation of second degree we get

$a=6,2h=5,b=-4,2g=7,2f=m,c=2$

Putting these values in $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$ gives us $m=2$ or $\frac{23}{6}$.

$a=4,2h=24,b=11$ from the given equation.

Let $\theta$ be the angle between the pair of straight lines. Then

$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}=\frac{4}{3}$, so the acute angle between the lines is ${\tan }^{-1}\frac{4}{3}$.

Given equation is $(x^2+y^2){\sin }^2\alpha =(x\cos \beta -y\sin \beta {)}^2$.

$\Rightarrow ({\cos }^2\beta -{\sin }^2\alpha )x^2-2\sin \beta \cos \beta xy+({\sin }^2\beta -{\sin }^2\alpha )y^2=0$

$\Rightarrow a={\cos }^2\beta -{\sin }^2\alpha ,2h=-\sin 2\beta ,b={\sin }^2\beta -{\sin }^2\alpha$

Let $\theta$ be the angle between the lines. Then

$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}=\pm \tan 2\alpha$

Thus, angle between the straight lines is $2\alpha$.

Comparing with the general equation, we have

$a=1,2h=-5,b=4$. If $\theta$ is the angle between the pair of straight lines, then

$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}=\frac{3}{5}$

Thus, acute angle between the lines is $ta{n}^{-1}\frac{3}{5}$.

We can write the given equation as $y^2-\frac{2}{\sin \theta \cos \theta }xy+\frac{{\sin }^4\theta +{\cos }^4\theta }{{\sin }^2\theta {\cos }^2\theta }{x}^2=0$

Let the straight lines by $y-x\tan \alpha =0$ and $y-x\tan \beta =0$, then

$\tan \alpha +\tan \beta =\frac{2}{\sin \theta \cos \theta }$, and $\tan \alpha \tan \beta =\frac{{\sin }^4\theta +{\cos }^4\theta }{{\sin }^2\theta {\cos }^2\theta }$

$(\tan \alpha -\tan \beta {)}^2=(\tan \alpha +\tan \beta {)}^2-4\tan \alpha \tan \beta$

$=\frac{4}{{\sin }^2\theta {\cos }^2\theta }-4\frac{{\sin }^4\theta +{\cos }^4\theta }{{\sin }^2\theta {\cos }^2\theta }$

$=\frac{4}{{\sin }^2\theta {\cos }^2\theta }(1-{\sin }^4\theta -{\cos }^4\theta )=8$

$\Rightarrow \tan \alpha -\tan \beta =2\sqrt{2}$.

Let $\theta$ be the angle between the pair of lines given by the equaiton $a{x}^2+2hxy+b{y}^2=0$. Let $A,B$ be the foot of perpendiculars from $P(p,q)$ on the two lines.

If we draw a circle with $OP$ as diameter then both $A$ and $B$ will be on tihs circle because diameter $OP$ will make right angle at any point on the circumference.

$\therefore \mathrm{\angle }OAB=\mathrm{\angle }OPB=\alpha$ (let)

From $\mathrm{△}OAB$, we have $\frac{OB}{\sin \alpha }=\frac{AB}{\sin \theta }$, and from $\mathrm{△}OPB$, we have $\frac{OP}{\sin {90}^{\circ }}=\frac{OB}{\sin \alpha }=\frac{AB}{\sin \theta }$

$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}=\frac{2\sqrt{h^2-ab}}{\sqrt{(a-b{)}^2+4h^2}}\Rightarrow AB=OP\sin \theta =\frac{2\sqrt{p^2+q^2}\sqrt{h^2-ab}}{\sqrt{(a-b{)}^2+4h^2}}$

Figure 10.1. 

From previous question $AB=\frac{2\sqrt{p^2+q^2}\sqrt{h^2-ab}}{\sqrt{(a-b{)}^2+4h^2}}$

$\therefore 2k=\frac{2\sqrt{p^2+q^2}\sqrt{h^2-ab}}{\sqrt{(a-b{)}^2+4h^2}}$

Hence, locus of point $P(p,q)$ is $(x^2+y^2)(h^2-ab)=k^2[(a-b{)}^2+4h^2]$.

Comparing with the general equation yields $a=1,b=-1$. Since $a+b=0$, the lines will be perpendicular to each other.

Like previour problem $a+b=0$, the lines will be perpendicular to each other.

Comparing with the general equation yields $a=1,b=1,h=-p$.

If $\theta$ is the angle between lines then $\tan \theta =\frac{2\sqrt{p^2-1}}{2}=\sqrt{p^2-1}$

$\Rightarrow \theta ={\tan }^{-1}\sqrt{p^2-1}$.

Comparing with the general equation yields $a=1,b=1,h=-\sec \theta$.

If $\alpha$ is the angle between the lines then $\tan \alpha =\frac{2\sqrt{{\sec }^2\theta -1}}{2}=\tan \theta$

Thus, the angle made by the pair of straight lines with one another is $\theta$.

The given equation is $(x^2+y^2){\sin }^2\alpha =(x\cos \alpha -y\sin \alpha {)}^2$.

Expanding the right-hand side gives $x^2{\cos }^2\alpha -2xy\sin \alpha \cos \alpha +y^2{\sin }^2\alpha$.

So the equation becomes $x^2{\sin }^2\alpha +y^2{\sin }^2\alpha =x^2{\cos }^2\alpha -2xy\sin \alpha \cos \alpha +y^2{\sin }^2\alpha$.

Canceling $y^2{\sin }^2\alpha$ from both sides gives $x^2{\sin }^2\alpha -x^2{\cos }^2\alpha +2xy\sin \alpha \cos \alpha =0$.

Using identities ${\sin }^2\alpha -{\cos }^2\alpha =-\cos 2\alpha$ and $2\sin \alpha \cos \alpha =\sin 2\alpha$, we get $-x^2\cos 2\alpha +xy\sin 2\alpha =0$.

Factoring gives $x(-x\cos 2\alpha +y\sin 2\alpha )=0$.

So the two lines are $x=0$ and $y\sin 2\alpha -x\cos 2\alpha =0$.

The first line has slope undefined, and the second line has slope $m=\frac{\cos 2\alpha }{\sin 2\alpha }=\cot 2\alpha$.

Using the angle formula between lines, the angle between them is $2\alpha$.

The given equation is $6x^2-5xy-6y^2+14x+5y+4=0$.

Compare with the general second degree form $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$.

We identify $a=6,2h=-5$ so $h=-\frac{5}{2}$, and $b=-6$.

To check if it represents a pair of straight lines, we use the condition $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$.

Here $g=7,f=5/2$, and $c=4$.

Substituting gives $6.(-6).4+2.\frac{5}{2}.7.(-\frac{5}{2})-6.{\left(\frac{5}{2}\right)}^2-(-6){.7}^2-4.{(-\frac{5}{2})}^2$.

This simplifies to $-144-\frac{175}{2}-\frac{150}{4}+294-\frac{100}{4}=0$, so the condition is satisfied.

Hence the equation represents a pair of straight lines.

Now for perpendicularity, we use the condition $a+b=0$.

Here $a+b=6+(-6)=0$, so the lines are perpendicular.

We identify $a=16,2h=24$ so $h=12$, and $b=9$.

First check if it represents a pair of straight lines using the condition $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$.

Here $g=20,f=15$, and $c=-75$.

Substituting gives $16.9.(-75)+2.15.20.12-{16.15}^2-{9.20}^2-(-75){.12}^2$.

This simplifies to $-10800+7200-3600-3600+10800=0$, so the condition is satisfied.

Hence, it represents a pair of straight lines.

Now check for parallel lines using the condition $h^2=ab$.

Here $h^2={12}^2=144$ and $ab=16.9=144$.

Since $h^2=ab$, the two lines are parallel.

$a=1,h=-\frac{5}{2},b=4,g=\frac{1}{2},f=1$, and $c=-2$.

Check the condition $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$.

Substituting gives $1.4.(-2)+2.1.\frac{1}{2}.(-\frac{5}{2})-{1.1}^2-4.{\left(\frac{1}{2}\right)}^2-(-2).{(-\frac{5}{2})}^2$.

This simplifies to $-8-\frac{5}{2}-1-1+25/2=0$, so the condition is satisfied.

Hence, it represents a pair of straight lines.

Now the angle between the lines is given by $\tan \theta =2\frac{\sqrt{h^2-ab}}{a+b}$.

Substitute $h^2=\frac{25}{4}$ and $ab=4$. So $h^2-ab=\frac{25}{4}-\frac{16}{4}=\frac{9}{4}$.

Then $\tan \theta =\frac{2.\frac{3}{2}}{5}=\frac{3}{5}\Rightarrow \theta =ta{n}^{-1}\frac{3}{5}$.

$a=12,h=\frac{7}{2},b=-p,g=-9,f=\frac{q}{2},c=6$.

For perpendicular lines: $a+b=0$ gives $12-p=0$ so $p=12$.

Using $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$ gives $q=1$.

$2x^2+3xy-2y^2=0\Rightarrow (2x-y)(x+2y)=0$, which gives two straight lines through origin and perpendicular to one another.

Given equation can be written as $2x^2-(y-9)x-(y^2+3y-10)=0$

Solving for $x$ gives $4x=y-9\pm (3y+1)\Rightarrow x-y+2=0$ and $2x+y+5=0$.

We can write the given equation as a quadratic equation in $x$ as follows:

$2x^2+(5y+6)x+(3y^2+7y+4)=0$

Solving for $x$ yields $4x=-(5y+6)\pm (y+2)\Rightarrow x+y+1=0$ and $2x+3y+4=0$.

Point of intersection is $(1,-2)$ and angle between them would be $ta{n}^{-1d})\frac{1}{5}$.

We can write the given equation as a quadratic equation in $x$ as follows:

$8x^2+(8y+26)x+2y^2+13y+15=0$

Solving for $x$ yields $16x=-(8y+26)\pm 14\Rightarrow 4x+2y+3=0$ and $2x+y+5=0$.

Thus, we have a pair of parallel straight lines.

Perpendicular distance between them is $\frac{|5-\frac{3}{2}|}{\sqrt{2^2+1}}=\frac{7}{2\sqrt{5}}$.

We can write the given equation as a quadratic equation in $x$ as follows:

$x^2-(5y-1)x+(4y^2+2y-2)=0$

Solving for $x$ yields $2x=(5y-1)\pm 3(y-1)\Rightarrow x-4y+2=0$ and $x-y-1=0$

Lines parallel to these lines and passing through $(1,1)$ are given byy

$x-4y-(1-4.2)=0$ and $x-y-(1-1)=0$ i.e. $x-4y+3=0$ and $x=y$.

The combined equation is $(x-4y+3)(x-y)=0\Rightarrow x^2-5xy+4y^2+3x-3y=0$

Thus, angle between lines is $\tan \theta =\left|\frac{2\sqrt{\frac{25}{4}-4}}{1+4}\right|=\frac{3}{5}\Rightarrow \theta ={\tan }^{-1}\frac{3}{5}$.

Given pair of lines is $2x^2-5xy+2y^2=0\Rightarrow (x-2y)(2x-y)=0\Rightarrow x-2y=0$ and $2x-y=0$ are the two lines represenetd by it.

Figure 10.2. 

Let these sides are $OA$ and $OB$. Let the given diagonal be $AB$ i.e. $5x+2y=1$.

Solving with the parallel sides we get $B$ as $(\frac{1}{6},\frac{1}{12})$ and $A$ as $(\frac{1}{9},\frac{2}{9})$. Thus, mid-point is $(\frac{5}{36},\frac{11}{72})$.

Thus, equation of other diagonal would be $11x-10y=0$ as it passes through $H$ and $O$.

Thus, ${\mathrm{\Delta }}_{OAB}=\frac{1}{72}$ and area of parallelogram is $\frac{1}{36}$.

The given pair of lines is $y^2-8xy-9x^2=0\Rightarrow y-9x=0$ and $x+y=0$.

Figure 10.3. 

Let the vertex be $(\alpha ,\beta )$ then $9y+x-(\alpha +9\beta )=0$ and $x-y+(\alpha -\beta )=0$ will be the equation of two sides perpendicular to the bisectors.

Let two other vertices be $(x_1,y_1)$ and $(x_2,y_2)$. Then $9y_1+x_1=\alpha +9\beta$ and $x_2-y_2=\alpha -\beta$

Also, mid-points will be $(\frac{\alpha +x_1}{2},\frac{\beta +y_1}{2})$ and $(\frac{\alpha +x_2}{2},\frac{\beta +y_2}{2})$.

Solving we get $(x_1,y_1)=(\frac{9\beta -40\alpha }{41},\frac{40\beta +9\alpha }{41})$ and $(x_2,y_2)=(-\beta ,-\alpha )$

$P$ is collinear with these two points, therefore,

$\left|\begin{array}{ccc}f& g& 1\\ -\beta & -\alpha & \\ \frac{9\beta -40\alpha }{41}& \frac{40\beta +9\alpha }{41}& 1\end{array}\right|=0$

Thus, we get locus as $4(x^2+y^2)+(4g+5f)x+(4f-5g)y=0$.

The two pair of straight lines are $y-mx=\pm a\sqrt{1+m^2}$ and $y-nx=\pm a\sqrt{1+n^2}$

Clearly the two pairs are parallel to each other. The distance between the lines are $|2a|$ for all pairs.

Thus, the given pair of straight lines form a rhombus.

Comparing given equation with the general equation of second degree we have

$a=0,b=0,c=c,h=h,g=g,f=f$

Since the equation represents two straight lines, therefore,

$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0\Rightarrow c=\frac{2fg}{h}$

Thus, the equation becomes $2hxy+2gx+2fy++\frac{2fg}{h}=0\Rightarrow (hx+f)(hy+g)=0$

$x=-\frac{f}{h},y=-\frac{g}{h}$, and clearly, these lines will make a rectangle with coordinate axes.

Area of the rectangle is $|-\frac{f}{h}|\times |-\frac{g}{h}|=\frac{|fg|}{h^2}$.

Assume given line represents a pair of straight lines: $(x+Ay+B)(x+Cy+D)=0$.

Expanding, $x^2+(A+C)xy+AC{y}^2+(B+D)x+(AD+BC)y+BD=0$.

Comparing coefficients: $A+C=-5,AC=4,B+D=1,AD+BC=2,BD=-2$.

From $AC=4$ and $A+C=-5$, we get $A=-1,C=-4$.

Now solve for $B,D$ using $B+D=1$ and $BD=-2$, giving $B=-1,D=2$.

Thus the factorization is $(x-y-1)(x-4y+2)=0$.

Hence, the separate equations of the lines are $x-y-1=0$ and $x-4y+2=0$.

The given homogeneous equations represent pairs of straight lines through the origin.

First equation is $x^2-6xy+3y^2=0$. Divide by $x^2$ assuming $x\ne 0$ to get $1-6\frac{y}{x}+3{\left(\frac{y}{x}\right)}^2=0$.

Let $m=\frac{y}{x}$. Then $3m^2-6m+1=0$.

The slopes are given by $m=\frac{6\pm \sqrt{36-12}}{6}=\frac{6\pm 2\sqrt{6}}{6}=1\pm \sqrt{6}/3$.

So the slopes are $m_1=1+\frac{\sqrt{6}}{3}$ and $m_2=1-\frac{\sqrt{6}}{3}$.

Second equation is $3x^2+6xy+y^2=0$. Divide by $x^2$ to get $3+6\frac{y}{x}+{\left(\frac{y}{x}\right)}^2=0$.

Let $m=\frac{y}{x}$. Then $m^2+6m+3=0$.

The slopes are $m=\frac{-6\pm \sqrt{36-12}}{2}=-3\pm \sqrt{6}$.

So the slopes are $m_3=-3+\sqrt{6}$ and $m_4=-3-\sqrt{6}$.

Now check perpendicularity using the condition that the product of slopes is $-1$.

Compute $m_1{m}_3=(1+\frac{\sqrt{6}}{3})(-3+\sqrt{6})=-3+\sqrt{6}-\sqrt{6}+2=-1$.

So $m_1$ is perpendicular to $m_3$.

Compute $m_2{m}_4=(1-\frac{\sqrt{6}}{3})(-3-\sqrt{6})$ $=-3-\sqrt{6}+\sqrt{6}+2=-1$.

So $m_2$ is perpendicular to $m_4$.

Thus, each line from the first equation is perpendicular to one line from the second equation.

Given pairs of lines are $y^2+xy-12x^2=0$ and $4y^2-13xy+3x^2=0$

$y^2+xy-12x^2=y^2+4xy-3xy-12x^2=0\Rightarrow (y+4x)(y-3x)=0$

$4y^2-13xy+3x^2=4y^2-12xy-xy+3x^2=(y-3x)(4y-x)=0$

We see that $y-3x=0$ conincides and the other pair i.e. $y+4x=0$ and $4y-x=0$ are perpendicular to one another.

$x^2-7xy+12y^2=x^2-3xy-4xy+12y^2=(x-3y)(x-4y)=0$

$12x^2+7xy+y^2=12x^2+3xy+4xy+y^2=(4x+y)(3x+y)=0$

Clearly, the lines are perpendicular to each other.

From $x^2-7x+6=0$ we get $(x-1)(x-6)=0$, so the lines are $x=1$ and $x=6$.

From $y^2-14y+40=0$ we get $(y-4)(y-10)=0$, so the lines are $y=4$ and $y=10$.

These four lines form a rectangle with vertices at $(1,4),(1,10),(6,4)$ and $(6,10)$.

The diagonals join opposite vertices.

First diagonal passes through $(1,4)$ and $(6,10)$. Its slope is $\frac{10-4}{6-1}=\frac{6}{5}$.

Using point-slope form $y-4=\frac{6}{5}(x-1)$ which simplifies to $6x-5y+14=0$.

Second diagonal passes through $(1,10)$ and $(6,4)$. Its slope is $\frac{4-10}{6-1}=-\frac{6}{5}$.

Using point-slope form $y-10=(-\frac{6}{5})(x-1)$ which simplifies to $6x+5y-56=0$.

The equation $3x^2+8xy-3y^2=0$ represents a pair of straight lines through the origin.

Divide by $x^2$ and put $m=y/x$. Then $3+8m-3m^2=0$ which gives $3m^2-8m-3=0$.

Solving, $m=\frac{8\pm \sqrt{64+36}}{6}$. So $m=3$ and $m=-\frac{1}{3}$.

Thus the lines are $y=3x$ and $y=-\frac{x}{3}$.

Now consider $3x^2+8xy-3y^2+2x-4y-1=0$. This represents another pair of lines parallel to the first pair.

Rewrite it as $3x^2+8xy-3y^2+2x-4y-1=(y-3x+1)(3y+x+1)$.

So the lines are $y=3x-1$ and $y=-\frac{x}{3}-1/3$.

Hence, we have two pairs of parallel lines $y=3x$ and $y=3x-1,y=-\frac{x}{3}$ and $y=-\frac{x}{3}-\frac{1}{3}$.

The slopes satisfy $3.(-\frac{1}{3})=-1$, so the adjacent sides are perpendicular. Also the perpendicular distance between each pair is equal.

Distance between $y=3x$ and $y=3x-1$ is $\frac{1}{\sqrt{1+9}}=\frac{1}{\sqrt{10}}$.

Distance between $y=-\frac{x}{3}$ and $y=-\frac{x}{3}-\frac{1}{3}$ is $\frac{\frac{1}{3}}{\sqrt{1+\frac{1}{9}}}=\frac{1}{\sqrt{10}}$.

Thus the four lines form a square.

Vertices are obtained by intersection of adjacent lines

Intersection of $y=3x$ and $y=-\frac{x}{3}$ gives $(0,0)$. Intersection of $y=3x-1$ and $y=-\frac{x}{3}$ gives $(\frac{3}{10},-\frac{1}{10})$. Intersection of $y=3x$ and $y=-\frac{x}{3}-\frac{1}{3}$ gives $(-\frac{1}{10},-\frac{3}{10})$, and intersection of $y=3x-1$ and $y=-\frac{x}{3}-\frac{1}{3}$ gives $(\frac{1}{5},-\frac{2}{5})$.

One diagonal joins $(0,0)$ and $(\frac{1}{5},-\frac{2}{5})$. Slope is $-2$ so equation is $y=-2x$ or $2x+y=0$.

Other diagonal joins $(\frac{3}{10},-\frac{1}{10})$ and $(-\frac{1}{10},-\frac{3}{10})$. Slope is $\frac{1}{2}$. Using point-slope form $y+\frac{1}{10}=\frac{1}{2}(x-\frac{3}{10})$ which simplifies to $x-2y-\frac{1}{2}=0$.

$2x^2-5xy-3y^2-2x+6y=(2x^2-5xy-3y^2)+(-2x+6y)$.

Factor the quadratic part $2x^2-5xy-3y^2=(2x+y)(x-3y)$.

So the equation becomes $(2x+y)(x-3y)-2(x-3y)$.

Factor out $(x-3y)\Rightarrow (x-3y)(2x+y-2)$.

Hence, the given equation reduces to $(x-3y)(2x+y-2)=0$.

Therefore, it represents two straight lines $x-3y=0$ and $2x+y-2=0$.

From $x-3y=0$ we get $x=3y$. Substitute into $2x+y-2=0\Rightarrow 2(3y)+y-2=0\Rightarrow 7y=2$ so $y=2/7$. Then $x=3y=\frac{6}{7}$.

Let one line represented by $a{x}^2+2hxy+b{y}^2=0$ is $y-mx=0$. Thus,

$a{m}^2{x}^2+2hxmx+b{m}^2{x}^2=0\Rightarrow a+2hm+b{m}^2=0$

Then according to question on of the lines of $a_1{x}^2+2h_{1}xy+b_1{y}^2=0$ would be $my+x=0$

$\Rightarrow a_1(-my{)}^2+2h(-my)y+b_1{y}^2=0\Rightarrow b_1-2h_{1}m+a_1{m}^2=0$

From two obtained equation we cross-multiply to get

$\frac{1}{2(h{a}_1+h_{1}b)}=\frac{m}{b{b}_1-a{a}_1}=\frac{m62}{-2(a{h}_1+b_{1}h)}$

Thus, $\frac{b{b}_1-a{a}_1}{2(h{a}_1+h_{1}b)}=-\frac{2(a{h}_1+b-1h)}{b{b}_1-a{a}_1}$

$\Rightarrow (b{b}_1-a{a}_1{)}^2+4(h_{1}a+b_{1}h)(h{a}_1+b{h}_1)=0$.

Consider the equation $a{x}^2+2hxy+b{y}^2=0$ which represents two straight lines through the origin.

Let their slopes be $m_1$ and $m_2$. Put $y=mx$ so that $a+2hm+b{m}^2=0$ or $b{m}^2+2hm+a=0$.

Thus, $m_1$ and $m_2$ are the roots of $b{m}^2+2hm+a=0$.

So $m_1+m_2=-\frac{2h}{b}$ and $m_1{m}_2=\frac{a}{b}$.

Given that one slope is $\lambda$ times the other, let $m_1=\lambda m_2$.

Then $m_1+m_2=\lambda m_2+m_2=(1+\lambda )m_2$ and $m_1{m}_2=\lambda m_2^2$.

Using $m_1{m}_2=\frac{a}{b}$ we get $\lambda m_2^2=\frac{a}{b}$ so $m_2^2=\frac{a}{b}\lambda$.

Now using $m_1+m_2=-\frac{2h}{b}$ we get $(1+\lambda )m_2=-\frac{2h}{b}$.

$\Rightarrow {\left(\frac{1+\lambda }{2h}\right)}^2=\frac{a}{b}\lambda$.

Consider the equation $a{x}^2+2hxy+b{y}^2=0$ which represents two straight lines through the origin.

Let their slopes be $m_1$ and $m_2$. Putting $y=mx$, we get $a+2hm+b{m}^2=0$ or $b{m}^2+2hm+a=0$.

Thus, $m_1$ and $m_2$ are the roots of this equation, so $m_1+m_2=-\frac{2h}{b}$ and $m_1{m}_2=\frac{a}{b}$.

Given that one slope is the square of the other, let $m_1=m_2^2$.

Then $m_1{m}_2=m_2^3=\frac{a}{b}$ so $m_2^3=\frac{a}{b}$.

Also $m_1+m_2=m_2^2+m_2=-\frac{2h}{b}$.

Multiply this by $m_2,m_2^3+m_2^2=(-\frac{2h}{b})m_2$.

Using $m_2^3=\frac{a}{b}$, we get $\frac{a}{b}+m_2^2=(-\frac{2h}{b})m_2$.

Multiply by $b,a+b{m}_2^2=-2h{m}_2$.

Now square both sides, ${(a+b{m}_2^2)}^2=4h^2{m}_2^2$. Expand, $a^2+2ab{m}_2^2+b^2{m}_2^4=4h^2{m}_2^2$.

Divide throughout by $ab,\frac{a}{b}+2m_2^2+\frac{b}{a}{m}_2^4=\frac{4h^2}{ab}{m}_2^2$.

Now use $m_2^3=\frac{a}{b}$. Then $m_2^4=m_2.m_2^3=m_2.\frac{a}{b}$ and $m_2^2=\frac{\frac{a}{b}}{m_2}$.

Substitute these into the equation and simplify. After simplification, we obtain $\frac{a+b}{h}+\frac{8h^2}{ab}=6$.

Each represents a pair of straight lines through the origin. Let a common line have slope $m$. Then it must satisfy both equations.

Putting $y=mx$ in each equation, we get $a+2hm+b{m}^2=0$ and $a^{\prime }+2h^{\prime }m+b^{\prime }{m}^2=0$.

Thus $m$ is a common root of the two quadratic equations $b{m}^2+2hm+a=0$ and $b^{\prime }{m}^2+2h^{\prime }m+a^{\prime }=0$.

For these two quadratics to have a common root, the condition is $(a{b}^{\prime }-a^{\prime }b{)}^2=4(a{h}^{\prime }-a^{\prime }h)(h{b}^{\prime }-h^{\prime }b)$.

Comparing given equation with the general equation in second degree gives us

$a=3,2h=-5$ and $b=4$.

We know that the equation of bisector of angles is given by $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

Substituting the values yields $5x^2-2xy-5y^2=0$.

The line $x+2y=7$ will be equally inclined to the lines $135x^2-136xy+33y^2=0$ if it is parallel to bisector of lines of $135x^2-136xy+33y^2=0$.

Comparing given equation with the general equation in second degree gives us

$a=135,2h=-136$ and $b=33$.

Equation of bisectors is given by $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

Substituting the values we obtain $2x^2+3xy-2y^2=0\Rightarrow (x+2y)(2x-y)=0$.

Clearly, the given line is parallel to $x+2y=0$.

The two pair of straight lines will be equally inclined if they have the same bisectors.

For $a{x}^2+2hxy+b{y}^2=0$ the bisectors are given by $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

For $a^2{x}^2+2h(a+b)xy+b^2{y}^2=0$ the bisectors are given by $\frac{x^2-y^2}{a^2-b^2}=\frac{xy}{h(a+b)}\Rightarrow \frac{x^2-y^2}{a-b}=\frac{xy}{h}$.

Thus, bisectors are same for them, and hence, they are equally inclined.

The bisectors of $a{x}^2+2hxy+b{y}^2=0$ are given by $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$.

$\Rightarrow h{x}^2-(a-b)xy-h{y}^2=0$, whose bisectors are given by

$\frac{x^2-y^2}{2h}=\frac{xy}{-\frac{a-b}{2}}\Rightarrow (a-b)(x^2-y^2)+4hxy=0$.

Equation of the bisectors for the second equation is $\frac{x^2-y62}{1-(-1)}=\frac{xy}{-q}\Rightarrow q{x}^2+2xy-q{y}^2=0$, which is given to be the same as first equation.

Comparing coefficients $\frac{1}{q}=\frac{-2p}{2}=\frac{1}{q}\Rightarrow pq=-1$.

Given pair of lines is $a{x}^2+2hxy+b{y}^2=\lambda (x^2+y^2)$

The equation of bisectors is given by $\frac{x^2-y^2}{a-\lambda -b+\lambda }=\frac{xy}{h}$, which is independent of $\lambda$. Hence, proven.

Equation of bisectors for lines $a{x}^2+acxy+c{y}^2=0$ is given by

$\frac{x^2-y^2}{a-c}=\frac{xy}{\frac{ac}{2}}=\frac{2xy}{ac}$.

Equation of bisectors for lines $(3+\frac{1}{c})x^2+xy+(3+\frac{1}{a})y^2=0$ us given by

$\frac{x^2-y^2}{3+\frac{1}{c}-3-\frac{1}{a}}=\frac{xy}{\frac{1}{2}}\Rightarrow \frac{x^2-y^2}{a-c}=\frac{xy}{\frac{ac}{2}}=\frac{2xy}{ac}$, which is same as previous bisectors.

The lines will be equally inclined if their bisectors are parallel. The bisectors of $2x^2+6xy+y^2=0$ are given by

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

The bisectors of $4x^2+18xy+y^2=0$ are given by

$\frac{x^2-y^2}{4-1}=\frac{xy}{9}\Rightarrow 3(x^2-y^2)=xy$.

Since the bisectors are same the lines are equally inclined.

Since the rotation of lines balance each other the bisectors in new position will be same as bisectors of the original position.

Thus, equation of bisectors of angles between the lines $x^2-2pxy-y^2=0$ is given by

$\frac{x^2-y^2}{2}=-\frac{xy}{p}$.

Let the pair of lines be given by the homogeneous second degree equation $a{x}^2+2hxy+b{y}^2=0$.

Since one of these lines is the bisector of the angle between the coordinate axes, it must be either $y=x$ or $y=-x$, because these are the angle bisectors of the axes.

Substitute $y=x$ into the given equation $a{x}^2+2hxx+b{x}^2=0$ which simplifies to $(a+2h+b)x^2=0$.

For this to represent a line, the coefficient must be zero, so $a+b+2h=0$.

Now substitute $y=-x$ into the equation $a{x}^2+2hx(-x)+b{x}^2=0$

which simplifies to $(a-2h+b)x^2=0$. Thus we get $a+b-2h=0$.

In either case, the condition for one of the lines to be an angle bisector is that either $a+b+2h=0$ or $a+b-2h=0$.

Both can be written together as $a+b=\pm 2h$. Squaring both sides gives $(a+b{)}^2=4h^2$.

First consider the pair of lines $x^2+xy-2y^2+4x-y+3=0$.

Assume a line through the origin has slope $m$, so it is $y=mx$. Substituting in the homogeneous equation gives $x^2+m{x}^2-2m^2{x}^2=0$. (we consider only homogeneous part)

After dividing by $x^2$ we obtain $1+m-2m^2=0$.

Thusm the slopes of the two given lines satisfy $2m^2-m-1=0$.

Let the roots be $m_1$ and $m_2$. Then $m_1+m_2=\frac{1}{2}$ and $m_1{m}_2=-\frac{1}{2}$.

The slopes of the angle bisectors satisfy $(1-m_1{m}_2)m^2+(m_1+m_2)m-(1-m_1{m}_2)=0$.

Substituting the values gives $(1+\frac{1}{2})m^2+\frac{m}{2}-(1+\frac{1}{2})=0$.

Multiplying by $2$ simplifies this to $3m^2+m-3=0$.

Hence, the bisectors through the origin have slopes satisfying $3m^2+m-3=0$.

Now the required lines pass through $(1,2)$ and are parallel to these bisectors, so their form is $y-2=m(x-1)$.

Eliminating $m$ using the quadratic condition gives the combined equation $3(y-2{)}^2+(y-2)(x-1)-3(x-1{)}^2=0$.

Given line is $fx-gy=\lambda \Rightarrow \frac{fx-gy}{\lambda }=1$.

Given equation is $x^2+hxy-y^2+gx+fy=0$

$\Rightarrow x^2+hxy-y^2+gx\frac{fx-gy}{\lambda }+fy\frac{fx-gy}{\lambda }=0$

$\Rightarrow (\lambda +fg)x^2+(\lambda h-g^2+f^2)xy-(\lambda +fg)y^2=0$

The above equation represents two lines through origin to the point of intersection of given line and curve.

Coeff. of $x^2+$ Coeff. of $y^2=0$. Hence, the lines are perpendicular to each other.

$y-3x=2\Rightarrow \frac{y-3x}{2}=1$. Now equation of the curve is $x^2+2xy+3y^2+4x+8y-11=0$