Straight Lines and Pair of Straight Lines

Let the centroid be $G(x, y)$. The coordinates of the centroid of triangles with vertices $\left(x_1, y_1\right),\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ are

$ x=\frac{x_1+x_2+x_3}{3} \text { and } y=\frac{y_1+y_2+y_3}{3} $

Substituting the coordinates of $A, B$ and $P$.

$ \begin{aligned} & x=\frac{2+3+t}{3}=\frac{5+t}{3}, \\ & y=\frac{3+2+t^2}{3}=\frac{5+t^2}{3} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & 3 x-5=t, y=\frac{5+(3 x-5)^2}{3} \\ & (\because t=2 x-5) \\ \Rightarrow & 3 y=5+9 x^2+25-30 x \\ \Rightarrow & 9 x^2-30 x-3 y+30=0 \\ \Rightarrow & 3 x^2-10 x-y+10=0 \end{array} $

The equation of the locus of the centroid of $\triangle A B C$ is $3 x^2-10 x-y+10=0$

Let $S(x, y)=2 x^2-x y-y^2-3 x+3 y$

$ \frac{\partial S}{\partial x}=4 x-y-3, \frac{\partial S}{\partial y}=-x-2 y+3 $

Solve, $\frac{\partial S}{\partial x}=0$ and $\frac{\partial S}{\partial y}=0$, we get

$ 4 x-y-3=0 \text { and }-x-2 y+3=0 $

Solving these two equations for $x=h$ and $y=k$

$ \begin{aligned} & 4 h-k-3=0 \text { and }-h-2 k+3=0 \\ & \therefore h=1, k=1 \end{aligned} $

So, the new origin is $(h, k)=(1,1)$

We know that a general second-degree equation

$A x^2+B x y+C y^2+D x+E y+F=0$, the angle of rotation $\theta$ satisfies

$ \tan 2 \theta=\frac{B}{A-C} $

Here, $S \equiv 2 x^2-x y-y^2-3 x+3 y=0$, we have

$ \begin{aligned} & A=2 B=-1 \text { and } C=-1 \\ & \therefore \tan 2 \theta=\frac{-1}{2-(-1)}=-\frac{1}{3} \end{aligned} $

Now, $h+k=1+1=2 h-k=1-1=0$,

$ h k=1 \times 1=1, \frac{h}{3 k}=\frac{1}{3} $

So, $\tan 2 \theta=\frac{-1}{3}=\frac{-h}{3 k}$

Give line is $5 x-12 y+6=0$

Slope of this line is $m_1=\frac{-5}{-12}=\frac{5}{12}$

Since, line $L$ is perpendicular to this line, the product of their slope is -1 . Let the slope of line $L$ be $m_L$.

Then, $m_L \times \frac{5}{12}=-1$

$ \Rightarrow \quad m_L=-\frac{12}{5} $

The equation of a line in normal form is $x \cos \theta+y \sin \theta=P$, where $P$ is the distance from the origin to the line and $\theta$ is the angle made by the perpendicular from the origin.

We have $P=2$ unit, So, the equation of line is $x \cos \theta+y \sin \theta=2$

Slope, $m_L=\frac{-\cos \theta}{\sin \theta}=-\cot \theta$

$ \therefore-\cot \theta=\frac{-12}{5} \Rightarrow \cot \theta=\frac{12}{5} $

For $y$-intercept, $x=0$ then $y \sin \theta=2$

$ \Rightarrow \quad y=\frac{2}{\sin \theta} $

And since $y$-intercept is positive, so

$ \begin{aligned} & \frac{2}{\sin \theta}>0 \\ & \Rightarrow \sin \theta>0 \end{aligned} $

$\Rightarrow \cot \theta=\frac{12}{5}>0, \theta$ must be in first quadrant.

Now, $\tan \theta=\frac{1}{\cot \theta}=\frac{5}{12}$

So, $\tan \theta+\cot \theta=\frac{5}{12}+\frac{12}{5}$

$ =\frac{25+144}{60}=\frac{169}{60} $

Let the angle made by Line $L$ with positive $X$-axis be $\theta$.

So, $B=(A x+A B \cos \theta, A y+A B \sin \theta)$

Given, $A(2,3)$ and $A B=4$, we have

$ B=(2+4 \cos \theta, 3+4 \sin \theta) $

Since, point $B$ lies on the line $4 x-3 y-19=0$, substitute the coordinate of $B$ into this equation.

$ \begin{aligned} & 4(2+4 \cos \theta)-3(3+4 \sin \theta)-19=0 \\ & \Rightarrow \quad 8+16 \cos \theta-9-12 \sin \theta-19=0 \\ & \Rightarrow \quad 16 \cos \theta-12 \sin \theta-20=0 \\ & \Rightarrow \quad 4 \cos \theta-3 \sin \theta-5=0 \\ & \Rightarrow \quad \frac{4}{5} \cos \theta-\frac{3}{5} \sin \theta=1 \end{aligned} $

Let $\cos \alpha=\frac{4}{5}$ and $\sin \alpha=\frac{3}{5}$ for some angle $\alpha$ in the first quadrant.

$\therefore \cos \alpha \cdot \cos \theta-\sin \alpha \cdot \sin \theta=1$

$ \begin{array}{ll} \Rightarrow & \cos (\alpha+\theta)=1 \\ \Rightarrow & \alpha+\theta=2 n \pi, \text { for some integer } n . \\ \Rightarrow & \alpha+\theta=0 \\ \Rightarrow & \alpha=-\theta \text { or } \alpha+\theta=2 \pi \\ \Rightarrow & \theta=2 \pi-\alpha \end{array} $

The angle made by the line $L$ with positive $X$-axis in the anti-clockwise direction is taken as a positive angle, where

$ \cos \theta=\frac{4}{5} \text { and } \sin \theta=-\frac{3}{5} $

This angle lies in the fourth quadrant.

So, $\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{-3}{4}$

$ \Rightarrow \quad \theta=\tan ^{-1}\left(\frac{-3}{4}\right)=-\tan ^{-1}\left(\frac{3}{4}\right) $

Let the equation of the line $L$ be $\frac{x}{a}+\frac{y}{b}=1$, where $a$ and $b$ are the $x$ and $y$ intercepts, respectively. So, $O A=a$ and $O B=b, a>0, b>0$.

Here, the line passes through the point $(4,9)$, So,

$ \frac{4}{a}+\frac{9}{b}=1 $

We want to minimize $a+b$. So consider

$ \begin{aligned} (a+b)\left(\frac{4}{a}+\frac{9}{b}\right) & =4+\frac{9 a}{b}+\frac{4 b}{a}+9 \\ & =13+\frac{9 a}{b}+\frac{4 b}{a} \end{aligned} $

By AM-GM inequality, for positive numbers,

$ \begin{aligned} & \frac{9 a}{b}+\frac{4 b}{a} \geq \sqrt[2]{\frac{9 a}{b} \cdot \frac{4 b}{a}} \\ \Rightarrow \quad & 2 \sqrt{36}=2 \cdot 6=12 \end{aligned} $

So, $13+\frac{9 a}{b}+\frac{4 b}{a}=13+12=25$

But, since $(a+b)\left(\frac{4}{a}+\frac{9}{b}\right)=(a+b) \cdot 1$

$ =a+b $     (Using Eq. (i))

$\therefore$ The minimum value of $a+b$ is 25 and this occurs when $\frac{9 a}{b}=\frac{4 b}{a}$

$ \begin{aligned} \Rightarrow & & 9 a^2 & =4 b^2 \text { or } 3 a=2 b \quad(\because a, b>0) \\ \Rightarrow & & b & =\frac{3}{2} a \end{aligned} $

So, $\frac{4}{a}+\frac{9}{b}=1$

$ \begin{aligned} & \Rightarrow \frac{4}{a}+\frac{9}{\frac{3}{2} a}=1 \Rightarrow \frac{4}{a}+\frac{18}{3 a}=1 \\ & \Rightarrow \frac{4}{a}+\frac{6}{a}=1 \Rightarrow \frac{10}{a}=1 \\ & \Rightarrow \quad a=10 \end{aligned} $

Then, $b=\frac{3}{2} a=\frac{3}{2} \times 10=15$

So, the minimum value of $O A+O B$ is $a+b=10+15=25$

Given equation,

$ 4 x^2+12 x y+9 y^2+2 g x+2 f y-1=0 $

Since, general second-degree equation $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ represents a pair of straight lines, if

$ \Delta=a b c+2 f g h-b g^2-c h^2-a f^2=0 $

Here, $a=4, b=9, h=6$

So, $\Delta=4 \times 9 \times c+2 f g \times 6-9 g^2$

$ \begin{aligned} & -c(6)^2-4 f^2=0 \\ & =36 c+12 f g-9 g^2-36 c-4 f^2=0 \\ & =12 f g-9 g^2-4 f^2=0 \\ \Rightarrow \quad & -(3 g-2 f)^2=0 \Rightarrow 3 g-2 f=0 \\ \Rightarrow \frac{g}{f} & =\frac{2}{3} \text { and } \frac{f}{g}=\frac{3}{2} \end{aligned} $

So, $\frac{f}{g}+\frac{g}{f}=\frac{3}{2}+\frac{2}{3}=\frac{9+4}{6}=\frac{13}{6}$

$x^2+3 x y-2 y^2+4 x-y-20=0$ substitute $x \rightarrow x^{\prime}+2, y \rightarrow y^{\prime}+3$ then,

$ \begin{aligned} & \left(x^{\prime}+2\right)^2+3\left(x^{\prime}+2\right)\left(y^{\prime}+3\right)-21\left(y^{\prime}+3\right)^2 \\ & +4\left(x^{\prime}+2\right)-\left(y^{\prime}+3\right)-20=0 \\ & \Rightarrow x^{\prime 2}+3 x^{\prime} y^{\prime}-2 y^{\prime 2}+17 x^{\prime}-7 y^{\prime}-11=0 \end{aligned} $

By comparing

$ A x^2+B x y+C y^2+D x+E y+F=0 $

We get $A=1, B=-3, C=-2, D=17$,

$ E=-7, F=-11 $

Therefore, $D+E+F=17-7-11=-1$

$L: 5 x-6 y+k=0$

$(2,3)$ and $(-4,-4 / 3)$ lie on opposite sides of the line for two points to be on opposite sides, the product at these points must be negative.

So, at $(2,3)$

$ \begin{aligned} & 5 \times 2-6 \times 3+k=k-8 \\ & \text { at }(-4,-4 / 3) \\ & 5(-4)-6(-4 / 3)+k=k-12 \end{aligned} $

Thus, $(k-8)(k-12)<0$

this means, $8

possible $k=9,10,11$

Similarly, $(1,2)$ and $(4,5)$ lies on the same side

$\therefore$ at $(1,2): 5 \times 1-6 \times 2+k=k-7$

and at $(4,5): 5 \times 4-6 \times 5+k=k-10$

Now, $(k-7)(k-10)>0$

This means, $k<7$ or $k>10$

that means, $k=11$ satisfy both conditions

$\therefore$ The equation of the line is

$ 5 x-6 y+11=0 $

Therefore, the perpendicular distance from the origin $(0,0)$ to the line $5 x-6 y+11=0$ is

$ \frac{11}{\sqrt{5^2+(-6)^2}}=\frac{11}{\sqrt{61}} $

Lines are : $x-2=0, x+y-1=0$, $x-y+3=0$ for vertices put $x=2$ in $x+y-1=0 \Rightarrow y=-1$ so, point is $(2,-1)$ Similarly, other vertices are $(2,5)$ and ( $-1,2$ ) Thus, side length will be $3 \sqrt{2}, 3 \sqrt{2}, 6$

Therefore,

Incentre

$ \begin{aligned} (\alpha, \beta) & =\left(\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}\right) \\ & =\left(\frac{3 \sqrt{2}(2)+3 \sqrt{2}(6)+6(-1)}{3 \sqrt{2}+3 \sqrt{2}+6}\right. , \left.\frac{3 \sqrt{2}(-1)+3 \sqrt{2}(5)+6(2)}{3 \sqrt{2}+3 \sqrt{2}+6}\right) \\ & =\left(\frac{4 \sqrt{2}-1}{\sqrt{2}+1}, 2\right) \end{aligned} $

Let the slopes of the lines represented by $3 x^2-4 x y+5 y^2=0$ be $m_1$ and $m_2$

then, $m_1+m_2=-\frac{(-4)}{3}$ and $m_1 m_2=5 / 3$

$\because$ Second pairs are perpendicular to the first pair.

then, slopes are $-\frac{1}{m_1}$ and $-\frac{1}{m_2}$ the given pair of lines are

$ l(x-a)^2+2 h(x-a)(y-b)+m(y-b)^2=0 $

then, $-\frac{1}{m_1}-\frac{1}{m_2}=\frac{-2 h}{l} \Rightarrow \frac{m_1+m_2}{m_1 m_2}=\frac{2 h}{l}$

and $\left(-\frac{1}{m_1}\right)\left(-\frac{1}{m_2}\right)=\frac{m}{l} \Rightarrow \frac{1}{m_1 m_2}=\frac{m}{l}$

Substitute the values of $m_1+m_2$ and $m_1 m_2$, we get

$ \begin{aligned} & -\frac{4 / 3}{5 / 3}=-\frac{2 h}{l} \\ \Rightarrow \quad & \frac{2 h}{l}=4 / 5 \Rightarrow 2 h=\frac{4}{5} l \end{aligned} $

and $\frac{1}{(5 / 3)}=m / l \Rightarrow \frac{m}{l}=3 / 5 \Rightarrow m=\frac{3}{5} l$

Thus, second pair of lines are

$ \begin{aligned} & l x^2+\frac{4}{5} l x y+\frac{3}{5} l y^2-32 x-26 y+c=0 \\ & \Rightarrow 5 x^2+4 x y+3 y^2-\frac{160}{l} x-\frac{130}{l} y+\frac{5 c}{l}=0 \end{aligned} $

Comparing with

$ l x^2+2 n x y+m y^2-32 x-26 y+c=0 $

We get $l=5, n=2, m=3$

Now, let the intersection point of second pair of line is $(a, b)$

The intersection of the lines is given by the equations is

$ 5 x+2 y=16 \text { and } 2 x+3 y=13 $

By solving these lines

We get $(x, y)=(a, b)=(2,3)$ and $c=71$

Hence, $\frac{a+b+c}{l+n+m}=\frac{2+3+71}{5+2+3}=\frac{76}{10}=\frac{38}{5}$

$\because \triangle P Q R$ is right angled at $P$ and $P Q=P R$

The slope of $Q R=-1$

So, the slope of angle bisector

$ =-\left(\frac{1}{-2}\right)=1 / 2 $

The equation of the angle bisector passing through $(2,1)$ is

$ y-1=\frac{1}{2}(x-2) \Rightarrow y=1 / 2 x $

$\because$ The angle bisector makes $45^{\circ}$ with each log.

Let the slope of one log be $m$ then, $\tan 45^{\circ}=\left|\frac{m-1 / 2}{1+m / 2}\right|$

$ \Rightarrow \quad \frac{m-1 / 2}{1+m / 2}= \pm 1 $

Thus, $m=3$ or $-1 / 3$

Therefore, the equation of $P Q(m=3)$

$ (y-1)=3(x-2) \Rightarrow 3 x-y-5=0 $

and the equation of $P R$

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

The pair of lines $P Q$ and $P R$

$ \begin{aligned} & (5 x-y-5)(x+3 y-5)=0 \\ & \Rightarrow 3 x^2+8 x y-3 y^2-20 x-10 y+25=0 \end{aligned} $

Rotation of Axes

$ \begin{array}{lllc} \hline \text { (Old) } & X & Y & \text { (New) } \\ \hline X & \cos \theta & -\sin \theta & X=X \cos \theta-Y \sin \theta \\ \hline y & \sin \theta & \cos \theta & y=X \sin \theta+Y \cos \theta \\ \hline \end{array} $

$ \begin{aligned} \because \quad \theta & =60^{\circ} \\ x & =\frac{X-\sqrt{3} Y}{2}, y=\frac{\sqrt{3} X+Y}{2} \\ \because x+y & =1 \end{aligned} $

$ \begin{gathered} X-\sqrt{3} Y+\sqrt{3} X+Y=2 \\ \Rightarrow X(1+\sqrt{3})+Y(1-\sqrt{3})=2 \\ a=\frac{2}{1+\sqrt{3}} \\ b=\frac{2}{1-\sqrt{3}} \\ \therefore \quad \frac{1}{a^2}+\frac{1}{b^2} \\ =\frac{(1+\sqrt{3})^2}{4}+\frac{(1-\sqrt{3})^2}{4} \\ \quad=\frac{1+3+1+3}{4}=\frac{8}{4}=2 \end{gathered} $

$ \begin{aligned} &\text { Let image of point }(2,-1) \text { w.r.t. line }\\ &\begin{aligned} & x-y+1=0 \text { be }(\alpha, \beta) \\ & \therefore \quad \frac{\alpha-2}{1}=\frac{\beta+1}{-1}=\frac{-2(2-(-1)+1)}{1^2+1^2} \\ & \Rightarrow \frac{\alpha-2}{1}=\frac{\beta+1}{-1}=-4 \\ & \Rightarrow \quad \alpha=-2, \beta=3 \\ & \therefore \text { Image }(-2,3) \end{aligned} \end{aligned} $

$ \text { Equation of line. } $

$ \begin{aligned} & \Rightarrow \quad x \cos \alpha+y \sin \alpha=P \\ & \Rightarrow \quad x \cos \frac{3 \pi}{4}+y \sin \frac{3 \pi}{4}=10 \\ & \Rightarrow \quad-x+y=10 \sqrt{2} \\ & \Rightarrow \quad x-y+10 \sqrt{2}=0 \end{aligned} $

Given, $3 x^2-2 y^2+a x y=0$

$ \begin{array}{lr} \Rightarrow & 2 y^2-a x y-3 x^2=0 \\ \Rightarrow & 2\left(\frac{y}{x}\right)^2-a\left(\frac{y}{x}\right)-3=0 \end{array} $

Equation has one root

$ \begin{array}{ll} \Rightarrow & m=\tan 60^{\circ}=\sqrt{3} \\ \therefore & 2(\sqrt{3})^2-a(\sqrt{3})-3=0 \\ \Rightarrow & \sqrt{3} a=3 \Rightarrow a=\sqrt{3} \end{array} $

Let coordinate of $P\left(\alpha, \frac{9-4 \alpha}{2}\right)$ and

$ Q(\beta,-6-2 \beta) $

$\because 0$ divides $P Q$ in $k: 1$

$ 0=\frac{k \beta+\alpha}{k+1}, 0=\frac{k(-6-2 \beta)+1\left(\frac{9-4 \alpha}{2}\right)}{k+1} $

$ \begin{aligned} \alpha & =-k \beta \\ k(6+2 \beta) & =\frac{9-4 \alpha}{2} \end{aligned} $

Put $\alpha=-k \beta$

$ \begin{array}{ll} \Rightarrow & 2 k(6+2 \beta)=9+4 k \beta \\ \Rightarrow & 12 k+4 k \beta=9+4 k \beta \\ \Rightarrow & k=\frac{9}{12} \\ \therefore & k=\frac{3}{4} \end{array} $

∴ Ratio $3: 4$

$ \text { Given, } A=(7,5), B(-5,-7) \text { and } C(7,-7) $

$\because \triangle A B C$ is right angled triangle

∴ Orthocentre is $M(7,-7)$

$ \begin{aligned} & \text { and incentre } \Rightarrow\binom{\frac{12 \times 7+12 \times-5+12 \sqrt{2}(7)}{24+12 \sqrt{2}},}{\frac{12 \times 5 \times 12(-7)+12 \sqrt{2}(-7)}{24+12 \sqrt{2})}} \\ & =\left(\frac{7-5+7 \sqrt{2}}{2+\sqrt{2}}, \frac{5-7-7 \sqrt{2}}{2+\sqrt{2}}\right)=\frac{2+7 \sqrt{2}}{2+\sqrt{2}}, \frac{-2-7 \sqrt{2}}{2+\sqrt{2}} \end{aligned} $

∴ Origin is shifted to $(7,-7) x_I{ }^{\prime}=x-h, y_I{ }^{\prime}=y-k$

$ \therefore \quad I=\left(\frac{2+7 \sqrt{2}}{2+\sqrt{2}}-7, \frac{-2-7 \sqrt{2}}{2+\sqrt{2}}+7\right)=\left(\frac{-12}{2+\sqrt{2}}, \frac{12}{2+\sqrt{2}}\right) $

$ \theta=\frac{\pi}{6}, b<0 $

Equation of line

$ \begin{aligned} & \Rightarrow \quad \frac{x}{-10}+\frac{y}{\frac{10}{\sqrt{3}}}=1 \Rightarrow \frac{x}{-10}+\frac{\sqrt{3} y}{10}=1 \\ & \Rightarrow \quad-x+\sqrt{3} y=10 \Rightarrow x-\sqrt{3} y-10=0 \end{aligned} $

Distance between $L$ and $(1,-\sqrt{3})$

$ D=\left|\frac{1+3-10}{\sqrt{4}}\right|=\frac{6}{2}=3 $

$m_{L_1}=2, m_{L_2}=\frac{-1}{2}$

Point of concurrency of $L_1$ and $L_2$

$ \begin{aligned} & x-y+2=0 \\ & 2 x+y+3=0 \end{aligned} $

On adding Eqs. (i) and (ii), we get

$ \begin{aligned} & 3 x=-5 \\ & x=-\frac{5}{3}, y=-\frac{5}{3}+2=\frac{1}{3} \end{aligned} $

∴ Point is $(-5 / 3,1 / 3)$

Equation of $L_1$

$ \begin{aligned} & \left(y-\frac{1}{3}\right)=2\left(x+\frac{5}{3}\right) \\ & \left(x_{\text {intercept }}\right)_{L_1}=\left|\frac{-1}{6} \frac{-5}{3}\right|=\frac{11}{6} \\ & \left(y_{\text {intercept }}\right)_{L_1}=\left|\frac{10}{3}+\frac{1}{3}\right|=\frac{11}{3} \end{aligned} $

Equation of $\boldsymbol{L}_{\mathbf{2}}$

$ \begin{aligned} & \left(y-\frac{1}{3}\right)=\frac{-1}{2}\left(x+\frac{5}{3}\right) \\ & \left(x_{\text {intercept }}\right)_{L_2}=\left(\frac{2}{3}-\frac{5}{3}\right)=1 \\ & \left(y_{\text {intercept }}\right)_{L_2}=\left|\frac{-5}{6}+\frac{1}{3}\right|=\frac{1}{2} \\ & +\frac{1}{2}=\frac{42}{6}=7 \end{aligned} $

∴ Required value $=\frac{11}{3}+\frac{11}{6}+1+\frac{1}{2}=\frac{42}{6}=7$

$ \begin{aligned} & \text { } L_1: y-x=0, L_3: y+2=0 \\ & \Rightarrow \quad P(-2,-2) \\ & \quad L_2: 2 x+y=0, L_3: y+2=0 \\ & \Rightarrow \quad Q(1,-2) \end{aligned} $

Bisector of $L_1$ and $L_2$

$ \frac{x-y}{\sqrt{2}}= \pm \frac{2 x+y}{\sqrt{5}} $

∵ Bisector divides in the ratio of sides

$ \frac{P R}{R Q}=\frac{O P}{O Q}=\frac{2 \sqrt{2}}{\sqrt{5}} $

Statement I is true and Statement II is false.

$ \begin{aligned} & \text { } 2 x^2+3 x y-2 y^2-5 x+2 f y-3=0 \ldots \text { (i) } \\ & a=2, b=-2, h=\frac{3}{2}, g=-\frac{5}{2}, f=f, c=-3 \end{aligned} $

∵ Eqn (i) represents a pair of straight lines

$ \begin{aligned} & \therefore \quad\left|\begin{array}{ccc} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} 2 & 3 / 2 & -5 / 2 \\ 3 / 2 & -2 & f \\ -5 / 2 & f & -3 \end{array}\right|=0 \\ & 2\left(6-f^2\right)-\frac{3}{2}\left(-\frac{9}{2}+\frac{5 f}{2}\right)-\frac{5}{2}\left(\frac{3 f}{2}-5\right)=0 \\ & \Rightarrow \quad 12-2 f^2+\frac{27}{4}-\frac{15 f}{4}-\frac{15 f}{4}+\frac{25}{2}=0 \\ & \Rightarrow \quad 48-8 f^2+27-15 f-15 f+50=0 \\ & \Rightarrow \quad 8 f^2+30 f-125=0 \\ & \Rightarrow \quad 8 f^2+50 f-20 f-125=0 \\ & \Rightarrow \quad 2 f(4 f+25)-5(4 f+25)=0 \\ & \Rightarrow \quad f=5 / 2 \frac{-25}{4} \end{aligned} $

Let the original coordinates of the point $P$ be $(x_0, y_0) = (4,1)$.

We will apply the transformations in succession and find the coordinates of $P$ after each step.

Step (i): Origin is shifted to the point $(1,6)$ by translation of axes.

When the origin is shifted to $(h,k)$, the new coordinates $(x',y')$ of a point $(x,y)$ are given by the formulas:

$x' = x - h$

$y' = y - k$

In this case, $(x,y) = (4,1)$ and $(h,k) = (1,6)$.

So, the coordinates of $P$ after this transformation, let's call them $(x_1, y_1)$, are:

$x_1 = 4 - 1 = 3$

$y_1 = 1 - 6 = -5$

The point $P$ is now at $(3,-5)$ with respect to the new coordinate system.

Step (ii): Translation through a distance of 2 units along the positive direction of X-axis.

This implies a translation of the point itself within the current coordinate system (established after step (i)). A translation of a point $(x,y)$ by $a$ units along the positive X-axis results in new coordinates $(x+a, y)$.

Here, the point is $(x_1, y_1) = (3,-5)$, and the translation distance is $a=2$.

So, the coordinates of $P$ after this transformation, let's call them $(x_2, y_2)$, are:

$x_2 = x_1 + 2 = 3 + 2 = 5$

$y_2 = y_1 = -5$

The point $P$ is now at $(5,-5)$.

Step (iii): Rotation of axes through an angle of $90^{\circ}$ in the positive direction.

This means the coordinate axes (with respect to which $P$ is currently at $(5,-5)$) are rotated by an angle $\theta = 90^{\circ}$ counter-clockwise.

If the old coordinates of a point are $(x,y)$ and the axes are rotated by an angle $\theta$, the new coordinates $(x',y')$ are given by:

$x' = x \cos\theta + y \sin\theta$

$y' = -x \sin\theta + y \cos\theta$

In this case, the point is $(x_2, y_2) = (5,-5)$ and $\theta = 90^{\circ}$.

$\cos 90^{\circ} = 0$

$\sin 90^{\circ} = 1$

So, the final coordinates of $P$, let's call them $(x_3, y_3)$, are:

$x_3 = 5 \cos 90^{\circ} + (-5) \sin 90^{\circ} = 5(0) + (-5)(1) = -5$

$y_3 = -5 \sin 90^{\circ} + (-5) \cos 90^{\circ} = -5(1) + (-5)(0) = -5$

The final position of the point $P$ is $(-5,-5)$.

Comparing this with the given options, the final coordinates are $(-5,-5)$, which corresponds to Option C.

The final answer is $\boxed{(-5,-5)}$.

Given, $L_1=a x-3 y+5=0$,

$ L_2=4 x-6 y+8=0 $

To find $x$-intercept of line $L_1$, set $y=0$, then

$ \begin{aligned} & & a x-3(0)+5 & =0 \\ \Rightarrow & & x & =-\frac{5}{a} \end{aligned} $

So, $x$-intercept $(p)=-\frac{5}{a}$

To find $y$-intercept $(q)$, set $x=0$, then

$ \begin{aligned} & a(0)-3 y+5 =0 \\ \Rightarrow & -3 y =-5 \end{aligned} $

So, $ q=y=\frac{5}{3} $

Now, to find $y$-intercept $(n)$, set $x=0$

So, $4(0)-6 y+8=0$

$ \Rightarrow \quad y=\frac{-8}{-6} $

So, $ n=y=\frac{4}{3} $

to find $x$-intercept $(m)$, set $y=0$ in equation $L_2$

$ \begin{aligned} & & 4(x)-6(0)+8 & =0 \\ \Rightarrow & & 4 x & =-8 \\ \therefore & & m & =x=-2 \end{aligned} $

Also, $L_1: a x-3 y+5=0$

$ \Rightarrow \quad y=\frac{a}{3} x+\frac{5}{3} $

$\therefore$ Slope of $L_1=\frac{a}{3}$

And $L_2: 4 x-6 y+8=0$

$ \Rightarrow \quad y=\frac{4}{6} x+\frac{8}{6}=\frac{2}{3} x+\frac{4}{3} $

$\therefore$ Slope of $L_2=\frac{2}{3}$

Since, $L_1$ and $L_2$ are parallel, so their slopes are equal.

$ \begin{array}{ll} \therefore & \frac{a}{3}=\frac{2}{3} \\ \Rightarrow & a=2 \end{array} $

Now, equation of the line passing through $(p, q)=\left(\frac{-5}{2}, \frac{5}{3}\right)$

And $(m, n)=\left(-2, \frac{4}{3}\right)$ is

$ \begin{array}{ll} & y-\frac{4}{3}=\frac{\frac{4}{3}-\frac{5}{3}}{-2-\left(\frac{-5}{2}\right)}(x-(-2)) \\ \Rightarrow & y-\frac{4}{3}=\frac{-1 / 3}{1 / 2}(x+2) \\ \Rightarrow & \frac{3 y-4}{3}=\frac{-2}{3}(x+2) \\ \Rightarrow & 3 y-4=-2 x-4 \\ \Rightarrow & 3 y+2 x=0 \end{array} $

The image of a point $\left(x_1, y_1\right)$ w.r.t to the line $a x+b y+c=0$ is

$ \begin{aligned} \frac{h-x_1}{a} & =\frac{k-y_1}{b} \\ & =\frac{-2\left(a x_1+b y_1+c\right)}{a^2+b^2} \end{aligned} $

Here $\left(x_1, y_1\right)=(2,-3)$, the line is $5 x-3 y-2=0$,

So, $a=5, b=-3$, and $c=-2$

Now, $\frac{h-2}{5}=\frac{k-(-3)}{-3}$

$ \begin{aligned} & =\frac{-2(5(2)+(-3)(-3)-2)}{5^2+(-3)^2} \\ \Rightarrow \quad \frac{h-2}{5} & =\frac{k+3}{-3} \\ & =\frac{-2(10+9-2)}{25+9} \\ \Rightarrow \quad \frac{h-2}{5} & =\frac{k+3}{-3}=\frac{-2(17)}{34}=-1 \end{aligned} $

So, $\frac{h-2}{5}=-1$ and $\frac{k+3}{-3}=-1$

$ \Rightarrow \quad h=-5+2=-3, k=3-3=0 $

So, $h+k=-3+0=-3$

Given equation is $2 x^2+x y-6 y^2=0$

$ \begin{aligned} & \Rightarrow \quad(2 x-3 y)(x+2 y)=0 \\ & \Rightarrow \quad 2 x-3 y=0 \text { and } x+2 y=0 \end{aligned} $

Now, $a x^2-7 x y-3 y^2=0$ and $2 x^2+x y-6 y^2$ has exactly one line in common.

So, either it has $2 x-3 y=0$ or $x+2 y=0$

Case $1 \,\,\mathrm{ax}^2-7 x y-3 y^2=0$

$ \begin{aligned} & \Rightarrow \quad(2 x-3 y)(p x+q y)=0 \\ & \Rightarrow \quad 2 p x^2+(2 q-3 p) x y-3 q y^2=0 \end{aligned} $

On comparing, we get

$ \begin{aligned} & 2 p=a, 2 q-3 p=-7,-3 q=-3 \\ \Rightarrow \quad & q=1 \end{aligned} $

So, $2(1)-3 p=-7$

$ \begin{array}{ll} \Rightarrow & 3 p=9 \\ \Rightarrow & p=3 \end{array} $

and $\quad 2 p=a$

$ \Rightarrow \quad a=2(3)=6 $

Case II $x+2 y=0$

So, $a x^2-7 x y-3 y^2=0$

$ \begin{aligned} & \Rightarrow \quad(x+2 y)(r x+s y)=0 \\ & \Rightarrow \quad r x^2+(s+2 r) x y+2 s y^2=0 \end{aligned} $

On comparing, we get

$ \begin{aligned} r & =a, s+2 r=-7,2 s=-3 \\ \Rightarrow \quad s & =\frac{-3}{2} \end{aligned} $

So, $s=+2 r=-7$

$ \begin{aligned} \Rightarrow 2 r & =-7-s \\ & =-7+\frac{3}{2}=-\frac{11}{2} \\ \Rightarrow r & =\frac{-11}{4} \end{aligned} $

So, $a=r=\frac{-11}{4}$

$ \therefore a=6, p=3, q=1 $

So, the lines are $(2 x-3 y)(3 x+y)=0$

$ \Rightarrow \quad 6 x^2-7 x y-3 y^2=0 $

Now, the bisectors are given by

$ \begin{aligned} & \frac{x^2-y^2}{6-(-3)}=\frac{x y}{-\frac{7}{2}} \\ \Rightarrow & \frac{x^2-y^2}{9}=\frac{2 x y}{-7} \\ \Rightarrow & -7 x^2+7 y^2=18 x y \\ \Rightarrow & 7 x^2+18 x y-7 y^2=0 \end{aligned} $

Given equation is

$ 2 x^2+2 h x y+2 y^2-x+y=1 $

The angle $\theta$ between the lines is $\tan ^{-1}\left(\frac{3}{4}\right)=0 \Rightarrow \tan \theta=\frac{3}{4}$. The angle $\theta$ between the pair of lines

$ \begin{aligned} & a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 \text { is } \\ & \tan \theta=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right| \end{aligned} $

Here $a=2, b=2$

$ \begin{aligned} \tan \theta & =\left|\frac{2 \sqrt{h^2-(2)(2)}}{2+2}\right|=\left|\frac{2 \sqrt{h^2-4}}{4}\right| \\ & =\left|\frac{\sqrt{h^2-4}}{2}\right| \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{3}{4}=\frac{\sqrt{h^2-4}}{2} \Rightarrow \frac{h^2-4}{4}=\frac{9}{16} \\ & \Rightarrow h^2-4=\frac{9}{16} \times 4=\frac{9}{4} \\ & \Rightarrow h^2=\frac{9}{4}+4=\frac{25}{4} \end{aligned} $

$ \Rightarrow h= \pm \sqrt{\frac{25}{4}}= \pm \frac{5}{2} $

Since, $h$ is a positive rational number,

$ h=\frac{5}{2} $

Thus, the given equation is

$ \begin{aligned} & 2 x^2+2 h x y+2 y^2-x+y-1=0 \\ & \quad 2 x^2+2 \cdot \frac{5}{2} x y+2 y^2-x+y-1=0 \\ & \quad 2 x^2+5 x y+2 y^2-x+y+1=0 \\ & \quad=F(x, y) \end{aligned} $

Now, $\frac{\partial F}{\partial x}=4 x+5 y-1=0$,

$ \frac{\partial F}{\partial y}=4 y+5 x+1=0 $

Solving these two equations, we get $x=-1, y=1$

So, the point of intersection is $(-1,1)$.

Locus of points which are equidistant from the coordinate are $y=x$ and $y=-x$

In $\triangle A O B$

Base $=A B=\sqrt{\left(3-(-3)^2+(3-3)^2\right.}=6$

and Height $(h)=3-0=3$

Therefore, area of $\triangle A O B=\frac{1}{2}(A B) \times h =\frac{1}{2} \times 6 \times 3=9$ sq. units

$4 x+3 y-1=0$

Normal vector $=\langle 4,3\rangle$

Direction vectors are perpendicular to the normal vector

$ \mathbf{d}=\langle 3,-4\rangle \text { or }\langle-3,4\rangle $

Unit vector in direction vector

$ \begin{aligned} & =\frac{1}{\sqrt{3^2+(-4)^2}}\langle 3,-4\rangle \\ & =\frac{1}{5}\langle 3,-4\rangle \end{aligned} $

So, two points 10 units away from $A=(-2,3)$ along the line are

$ \begin{aligned} & \left(x_1, y_1\right)=(-2,3)+10 \cdot \frac{1}{5}<3,-4> \\ & =(-2+6,3-8)=(4,-5) \\ & \text { and } \left.\left(x_2, y_2\right)=(-2,3)-10 \cdot \frac{1}{5}<3,-4\right) \\ & =(-2-6,3+8) \\ & =(-8,11) \end{aligned} $

Therefore, $\left(x_1+y_1\right)^2+\left(x_2+y_2\right)^2$

$ \begin{aligned} & =(4-5)^2+(-8+11)^2 \\ & =1+9=10 \end{aligned} $

$12 x-5 y+13=0$

$\therefore$ Normal vector $=\langle 12,-5\rangle$

Let $\alpha$ be the angle between the vector <12, -5>

and positive $X$-axis

So, $\tan \alpha=\frac{y}{x}=\frac{-5}{12}$

$ \begin{aligned} & \Rightarrow \quad \alpha=\tan ^{-1}\left(\frac{-5}{12}\right) \\ & \Rightarrow \quad \alpha=\pi-\tan ^{-1}\left(\frac{5}{12}\right) \end{aligned} $

$2 x^2+x y-y^2-x+2 y-1=0 \,\,\,\,\,\ldots$ (i)

Second degree (homogenous) part of Eq. (i)

$ 2 x^2+x y-y^2=0 $

assume $y=m x$, and substitute

$ 2 x^2+x(m x)-(m x)^2=0 $

Coefficient of $x^2$ should be zero

$ \begin{gathered} \Rightarrow \quad 2+m-m^2=0 \Rightarrow m=2,-1 \\ m_1=2 \text { and } m_2=-1 \end{gathered} $

$\because$ Both lines are perpendicular their slopes must be

$ -\frac{1}{m_1}=-\frac{1}{2} \text { and } \frac{-1}{m_2}=1 $

So, the perpendicular lines has their slopes $-1 / 2$ and 1

and passing through $(1,1)$

$ \begin{aligned} & \therefore \text { Pair of lines are } \\ & \left(y-y_1-m_1\left(x-x_1\right)\right) \\ & \left(y-y_1-m_2\left(x-x_2\right)\right)=0 \\ & \Rightarrow\left(y-1+\frac{1}{2}(x-1)\right)(y-1-(x-1))=0 \\ & \Rightarrow\left(\frac{x}{2}+y-\frac{3}{2}\right)(y-x)=0 \\ & \Rightarrow \frac{1}{2} x y-\frac{1}{2} x^2+y^2-x y-\frac{3}{2} y+\frac{3}{2} x=0 \\ & \Rightarrow-x^2-x y+2 y^2-3 y+3 x=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Compare with

$ \begin{gathered} a x^2+2 h x y+b y^2+2 g x+3 y=0 \\ \therefore \quad a=-1, h=+1 / 2, b=2 g=-3 / 2 \end{gathered} $

so, $\frac{b}{a}=\frac{2}{-1}=-2$

Put $x=2-y$ in the given conic

$ \begin{aligned} & \Rightarrow(2-y)^2+y^2-2(2-y)-4 y+2=0 \\ & \Rightarrow y^2-4 y+4+y^2-4+2 y-4 y+2=0 \\ & \Rightarrow 2 y^2-6 y+2=0 \Rightarrow y^2-3 y+1 \\ & \therefore y=\frac{3 \pm \sqrt{5}}{2} \\ & \Rightarrow y_1=\frac{3+\sqrt{5}}{2}, y_2=\frac{3-\sqrt{5}}{2} \end{aligned} $

Thus, $x_1=2-\left(\frac{3+\sqrt{5}}{2}\right)=\frac{1-\sqrt{5}}{2}$ and

$ x_2=2-\left(\frac{3-\sqrt{5}}{2}\right)=\frac{1+\sqrt{5}}{2} $

then, $p_1 \equiv\left(\frac{1-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right)$ and

$ p_2 \equiv\left(\frac{1+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}\right) $

The lines $O P_1$ and $O P_2$ will have equation of the form

$ \begin{aligned} & \quad \begin{array}{l} y=m x \\ \Rightarrow \quad l x+m y=0 \end{array} \\ & \quad \begin{aligned} \because \quad m_1 & =\frac{y_1}{x_1}=\frac{3+\sqrt{5}}{1-\sqrt{5}} \\ & =\frac{3+\sqrt{5}}{1-\sqrt{5}} \times \frac{1+\sqrt{5}}{1+\sqrt{5}}=-2-\sqrt{5} \end{aligned} \\ & \text { and } m_2=\frac{y_2}{x_2}=\frac{3-\sqrt{5}}{1+\sqrt{5}} \\ & \quad=\frac{3-\sqrt{5}}{1+\sqrt{5}} \times \frac{1-\sqrt{5}}{1-\sqrt{5}}=-2+\sqrt{5} \end{aligned} $

$\therefore$ Combined equation of lines are $\left(m_1 x-y_1\right)\left(m_2 x-y\right)=0$

will form $\Rightarrow\left(l_1 x+m_1 y\right)\left(l_2 x+m_2 y\right)=0$

$ \begin{aligned} &\text { Lets expands and compare }\\ &\begin{aligned} & m_1 m_2 x^2-\left(m_1\right.\left.+m_2\right) x y+y^2=l_1 l_2 x^2 \\ &+\left(l_1 m_2+l_2 m_1\right) x y+m_1 m_2 y^2 \\ & \therefore l_1 l_2=m_1 m_2, l_1 m_2+l_2 m_1=-\left(m_1+m_2\right), \\ & m_1 m_2=1 \\ & m_1=-2-\sqrt{5}, m_2=-2+\sqrt{5} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { and } m_1+m_2=-4 \\ & \because \quad l_1 l_2=m_1 m_2=1 \\ & \Rightarrow \quad l_1=1 / l_2 \\ & \therefore \quad l_1 m_2+\frac{m_1}{l_1}=4 \\ & \Rightarrow \quad\left(\sqrt{5}-2 l_1^2+(-\sqrt{5}-2)=4 l_1\right. \\ & \Rightarrow \quad l_1=1 \text { and } l_2=1 \\ & \text { Hence, } l_1+l_2+m_1+m_2=1+1-4=-2 \end{aligned} $

$ \text { Let } P \equiv(\alpha, \beta) $

$ \therefore \quad m_{A P}=\frac{\beta-4}{\alpha-5} $

$ \begin{aligned} &m_{P B}=\frac{\beta+4}{\alpha-5}\\ &\begin{aligned} & \text { Given, } \angle A P B=\frac{\pi}{4} \\ & \therefore \tan \frac{\pi}{4}=\left|\frac{\frac{\beta-4}{\alpha-5}-\frac{\beta+4}{\alpha-5}}{1+\frac{\beta-4}{\alpha-5} \times \frac{\beta+4}{\alpha-5}}\right| \\ & \Rightarrow\left|\frac{\frac{\beta-4-\beta-4}{\alpha-5}}{\frac{(\alpha-5)^2+\beta^2-16}{(\alpha-5)^2}}\right|=1 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{-8}{(\alpha-5)} \times \frac{(\alpha-5)^2}{(\alpha-5)^2+\beta^2-16}= \pm 1 \\ & \Rightarrow-8 \alpha+40=\alpha^2+25-10 \alpha+\beta^2-16 \\ & \text { or }-8 \alpha+40=-\left(\alpha^2+25-10 \alpha+\beta^2-16\right) \\ & \Rightarrow \alpha^2+\beta^2-2 \alpha-31=0 \\ & \text { or } \alpha^2+\beta^2-18 \alpha+49=0 \\ & =\text { Locus of } P \text { are } \\ & \quad x^2+y^2-2 x-31=0 \\ & \text { or } x^2+y^2-18 x+49=0 \end{aligned} $

$ \frac{|6+12-3|}{5}=\frac{|12+4 a-3|}{5} =\frac{|3 \alpha+4 \beta-3|}{5}$

      (i)                    (ii)                    (iii)

From (i) and (ii) we get

$ \begin{aligned} & |4 a+9|=15 \\ & \Rightarrow 4 a+9=15,-15 \\ & \Rightarrow \quad 4 a=6,-24 \Rightarrow a=\frac{3}{2},-6 \end{aligned} $

From (i) and (iii), we get

$ 3 \alpha+4 \beta-3= \pm 15 $

Case (I) When $3 \alpha+4 \beta-3=15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, $4 \alpha-3 \beta+1=0 \quad$ (given)                                 ...(ii)

On solving Eqs. (i) and (ii) we get

$ \alpha=2 \beta=3 $

Case (II) When $3 \alpha+4 \beta-3=-15$

$ \Rightarrow \quad 3 \alpha+4 \beta=-12 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

and $4 \alpha-3 \beta+1=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

On solving Eqs. (iii) and (iv) we get,

$ \alpha=\frac{-8}{5}, \beta=\frac{-9}{5} $

$\therefore$ Sum of all possible values of $a, \alpha, \beta=$

$ \begin{aligned} & \frac{3}{2}+2+3=\frac{3}{2}+5=\frac{13}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\\ & \text { or }-6-\frac{8}{5}-\frac{9}{5}=\frac{-30-17}{5}=\frac{-47}{5} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)\end{aligned} $

On adding Eqs. (v) and (vi)

$\therefore$ Total sum $=\frac{-29}{10}$

Equation of base of equilateral triangle is

$ \begin{gathered} x+y=2 \\ \therefore \text { Slope }=-1 \Rightarrow \tan 0=-1 \\ \theta=135^{\circ} \end{gathered} $

$ \begin{aligned} A D & =\left|\frac{2+1-2}{\sqrt{2}}\right|=\frac{1}{\sqrt{2}} \\ a & =A B=\frac{2}{\sqrt{3}} A D=\frac{2}{\sqrt{3}} \times \frac{1}{\sqrt{2}}=\sqrt{\frac{2}{3}} \\ a & =\sqrt{\frac{2}{3}} \\ \therefore \quad m_1 & =\tan \left(135^{\circ}-60^{\circ}\right) \\ \text { and } m_2 & =\tan \left(135^{\circ}-120^{\circ}\right) \\ m_1 & =\tan 75^{\circ} \text { and } m_2=\tan 15^{\circ} \\ m_1 & =2+\sqrt{3} \text { and } m_2=2-\sqrt{3} \\ \therefore\left|m_1-m_2\right| & =|2+\sqrt{3}-2+\sqrt{3}|=2 \sqrt{3} \\ \left|m_1-m_2\right|+a \sqrt{2} & =2 \sqrt{3}+\frac{\sqrt{2}}{\sqrt{3}} \times \sqrt{2} \\ & =\frac{6+2}{\sqrt{3}}=\frac{8}{\sqrt{3}} \end{aligned} $

$ \begin{aligned} & \text { } \quad 2 x^2+x y-6 y^2=0 \\ & \Rightarrow \quad 2 x^2+4 x y-3 x y-6 y^2=0 \\ & \Rightarrow \quad 2 x(x+2 y)-3 y(x+2 y)=0 \\ & \Rightarrow \quad(2 x-3 y)(x+2 y)=0 \\ & \Rightarrow \quad 2 x-3 y=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { or } \quad x+2 y=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Given, $x+y-1=0 \ldots$ (iii)

Now on solving Eqs. (i) and (ii) we gel. $(0,0)$.

On solving Eqs. (ii) and (iii), we get $(2,-1)$

and solving Eqs. (i) and (iii) we get,

$ \left(\frac{3}{5}, \frac{2}{5}\right) $

As we observe that, $A B \neq B C \neq A C$

Because, $A B=\sqrt{\frac{98}{25}}$

$\begin{aligned} B C & =\sqrt{5} \\ \text { and } \quad A C & =\sqrt{\frac{13}{25}}\end{aligned}$

$ \begin{aligned} &\text { Given, } 4 x^2-y^2=0\\ &\Rightarrow(2 x+y)(2 x-y)=0 \end{aligned} $

$ \begin{aligned} \Rightarrow \quad & 2 x-y=0 \\ & 2 x+y=0 \\ & l x+m y+n=0 \end{aligned} $

Consider $A B C$ is a triangle formed by the sides (i), (ii), (iii).

$ \begin{aligned} & \left.\begin{array}{c} A \equiv\left(\frac{-n}{l-2 m}, \frac{2 n}{l-2 m}\right) \\Now,\,\, B \equiv\left(\frac{-n}{l+2 m}, \frac{-2 n}{l+2 m}\right) \\ C \equiv(0,0) \\ \text { Given centroid }=\left(\frac{2}{3}, 0\right) \end{array}\right\} \end{aligned} $

(By solving Eqs. (i), (ii) and (iii) taken two eqns at a time.)

$ \begin{array}{ll} \therefore & \frac{-n}{l-2 m}-\frac{n}{l+2 m}+0=\frac{2}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ & \frac{2 n}{l-2 m}-\frac{2 n}{l+2 m}+0=0 \\ \Rightarrow & 2 n\left(\frac{1}{l-2 n}-\frac{1}{l+2 m}\right)=0 \\ \Rightarrow & n=0 \text { or } \frac{1}{l-2 m}=\frac{1}{l+2 m} \\ \Rightarrow & n=0 \text { or } l+2 m=l-2 m \\ \Rightarrow & m=0 \end{array} $

From Eq. (iv), $\frac{-n}{l-0}-\frac{n}{l+0}=\frac{2}{3}$

$ \begin{aligned} & \Rightarrow \quad \frac{-2 n}{l}=\frac{2}{3} \\ & \Rightarrow \quad l=-3 n \\ & \therefore l+m+n=-3 n+0+n=-2 n=0 \\ & \because(b c=n=0) \end{aligned} $

Step 1: Write the coordinates
Let $P$ be any point $(x, y)$. $A$ is $(1, 0)$. $B$ is $(0, -2)$. $C$ is $(2, -1)$.

Step 2: Find area of triangle $PAB$
The area of triangle $PAB$ is given by: $ \text{Area of }\triangle PAB = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| $ For $P(x, y), A(1, 0), B(0, -2)$: $ \text{Area} = \frac{1}{2}|x(0 + 2) + 1(-2 - y) + 0(y - 0)| = \frac{1}{2}|2x - 2 - y| $

Step 3: Find area of triangle $PAC$
Use the same formula for $P(x, y), A(1, 0), C(2, -1)$: $ \text{Area} = \frac{1}{2}|x(0 + 1) + 1(-1 - y) + 2(y - 0)| = \frac{1}{2}|x + y - 1| $

Step 4: Set the areas equal
We want these two areas to be the same: $ \frac{1}{2}|2x - 2 - y| = \frac{1}{2}|x + y - 1| $ Take both sides times $2$: $ |2x - 2 - y| = |x + y - 1| $

Step 5: Remove the modulus (absolute value)
Square both sides to get rid of absolute values: $ (2x - 2 - y)^2 = (x + y - 1)^2 $

Step 6: Expand both sides
Expand the left side: $ (2x - 2 - y)^2 = (2x - y - 2)^2 = 4x^2 + y^2 + 4 + 4xy - 8x - 4y $ Expand the right side: $ (x + y - 1)^2 = x^2 + y^2 + 1 + 2xy - 2x - 2y $

Step 7: Move all terms to one side
Subtract right side from left side: $ (4x^2 + y^2 + 4 + 4xy - 8x - 4y) - (x^2 + y^2 + 1 + 2xy - 2x - 2y) = 0 $ Simplify: $ 4x^2 - x^2 + 4xy - 2xy - 8x + 2x + (-4y + 2y) + 4 - 1 = 0 \\ 3x^2 + 2xy - 6x - 2y + 3 = 0 $

Step 8: Divide by 3
To make it simpler, divide all terms by 3: $ x^2 - 2xy - 2x + 2y + 1 = 0 $

$ \begin{aligned} &\text { } \tan \theta=2\\ &\begin{aligned} & \sin \theta=\frac{2}{\sqrt{5}} \\ & \cos \theta=\frac{1}{\sqrt{5}} \\ & x=x^{\prime} \cos \theta-y^{\prime} \sin \theta=\frac{x^{\prime}}{\sqrt{5}}-\frac{2 y^{\prime}}{\sqrt{5}} \\ & y=x^{\prime} \sin \theta+y^{\prime} \cos \theta=\frac{2 x^{\prime}}{\sqrt{5}}+\frac{y^{\prime}}{\sqrt{5}} \\ & \Rightarrow 3\left(\frac{x^{\prime}}{\sqrt{5}}-\frac{2 y^{\prime}}{\sqrt{5}}\right)^2-4\left(\frac{x^{\prime}}{\sqrt{5}}-\frac{2 y^{\prime}}{\sqrt{5}}\right)\left(\frac{2 x^{\prime}}{\sqrt{5}}+\frac{y^{\prime}}{\sqrt{5}}\right)=r^2 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{3 x^{\prime 2}-12 x^{\prime} y^{\prime}+12 y^{\prime 2}}{5} -\frac{8 x^{\prime 2}-12 x^{\prime} y^{\prime}-8 y^{\prime 2}}{5}=r^2 \\ & \Rightarrow-x^{\prime 2}+4 y^{\prime 2}=r^2 \end{aligned} $

$ \begin{aligned} &\Rightarrow \text { The transformed equation is }\\ &\,\,\,\,\,\,\,\,\,\,\,\,4 y^{\prime 2}-x^{\prime 2}=r^2 \end{aligned} $

$ \Rightarrow \quad 4 y^2-x^2=r^2 $

First, we need to find the point where the two lines $x-2y+3=0$ and $2x-y=0$ meet.

Step 1: Find Intersection Point
To solve the two equations:
$x - 2y + 3 = 0$
$2x - y = 0$

Use the second equation: $2x - y = 0 \implies y = 2x$.

Put $y = 2x$ into the first equation:

$x - 2(2x) + 3 = 0 \implies x - 4x + 3 = 0$
$-3x + 3 = 0$
$x = 1$

Substitute $x = 1$ back: $y = 2x = 2 \times 1 = 2$

So, the intersection is at the point $(1, 2)$.

Step 2: Find Another Point for Line $L_2$
$L_2$ passes through the origin (0,0) and the intersection of $3x - y + 2 = 0$ and $x - 3y - 2 = 0$.

We need the intersection point of $3x - y + 2 = 0$ and $x - 3y - 2 = 0$.

Multiply the second equation by $3$:
$3x - 9y - 6 = 0$

Subtract $3x - y + 2 = 0$ from $3x - 9y - 6 = 0$:
$[3x - 9y - 6] - [3x - y + 2] = 0$
$(3x - 9y - 6) - 3x + y - 2 = 0$
$-8y - 8 = 0$
$y = -1$

Substitute $y = -1$ into $x - 3y - 2 = 0$:
$x - 3(-1) - 2 = 0$
$x + 3 - 2 = 0$
$x = -1$

So, the intersection point is $(-1, -1)$.

Step 3: Find Equation of $L_2$
$L_2$ passes through $(0,0)$ and $(-1,-1)$.
Slope $m = \frac{-1-0}{-1-0} = 1$.

Using slope-intercept form: $y = x$.

In general form: $-x + y = 0$           ...(i)

Step 4: Find Equation of $L_1$
$L_1$ is parallel to $L_2$ (so same slope $m=1$) and passes through $(1,2)$.

Write $y = x + c$.
Plug in $(1,2)$:
$2 = 1 + c \implies c = 1$.

So Equation of $L_1$ is $y = x + 1$, or $-x + y = 1$           ...(ii)

Step 5: Find Distance Between $L_1$ and $L_2$
The distance between two parallel lines $-x + y = 1$ and $-x + y = 0$ is:

$\text{Distance} = \frac{|1 - 0|}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}$

$x+y-2=0$ and $3 x-4 y+1=0$

Solving these equation we get $x=1$, $y=1$

The intersection point is $(\alpha, \beta)=(1,1)$

$ \begin{aligned} & \Rightarrow 5 \cdot 1+k \cdot 1-7=0 \\ & k=2 \\ & \Rightarrow 2 x+y-2=0 \\ & \text { Slope } m_1=\frac{-2}{1}=-2 \\ & \qquad m_2=-\frac{1}{m_1}=-\frac{1}{-2}=\frac{1}{2} \\ & \qquad y-y_1=m\left(x-x_1\right) \\ & \Rightarrow \quad y-y_1=\frac{1}{2}\left(x-x_1\right) \\ & \Rightarrow \quad y-1=\frac{1}{2}(x-1) \\ & \therefore \quad x-2 y+1=0 \end{aligned} $

The equation $3x^2 - 5xy + 2y^2 = 0$ can be written as two factors: $(3x - 2y)(x - y) = 0$.

This means the two sides of the triangle are $3x - 2y = 0$ and $x - y = 0$.

The slope of $3x - 2y = 0$ is $m_1 = \frac{3}{2}$, and the slope of $x - y = 0$ is $m_2 = 1$.

To find where the triangle's heights (altitudes) go from the orthocentre at $(2,1)$, we need the slope of each altitude. The altitude to $3x-2y=0$ will have a slope that is the negative reciprocal of $\frac{3}{2}$, which is $-\frac{2}{3}$.

The altitude to $x-y=0$ will have a slope that is the negative reciprocal of $1$, which is $-1$.

Write the equation for the altitude with slope $-\frac{2}{3}$ that passes through $(2,1)$:

$ \begin{aligned} &y - 1 = -\frac{2}{3}(x - 2) \\ &\Rightarrow 3y - 3 = -2x + 4 \\ &\Rightarrow 2x + 3y = 7 \end{aligned} $

Write the equation for the altitude with slope $-1$ that passes through $(2,1)$:

$ \begin{aligned} &y - 1 = -1(x - 2) \\ &\Rightarrow y - 1 = -x + 2 \\ &\Rightarrow x + y - 3 = 0 \end{aligned} $

Find where each altitude meets the opposite side. First, where $2x + 3y = 7$ (altitude) meets $x-y=0$ (side):

Since $x = y$, substitute it into $2x + 3x = 7$, so $5x = 7$ which means $x = \frac{7}{5}$.

The intersection is at $\left(\frac{7}{5}, \frac{7}{5}\right)$.

Now, where $x + y - 3 = 0$ (altitude) meets $3x - 2y = 0$:

From $3x - 2y = 0$, $x = \frac{2}{3}y$.

Put this in $x + y - 3 = 0$ to get $\frac{2}{3}y + y = 3 \implies \frac{5}{3}y = 3 \implies y = \frac{9}{5}$.

When $y = \frac{9}{5}$, $x = \frac{6}{5}$, so the intersection point is $\left( \frac{6}{5}, \frac{9}{5} \right)$.

The third side of the triangle must be the line connecting these two intersection points: $\left(\frac{7}{5}, \frac{7}{5}\right)$ and $\left(\frac{6}{5}, \frac{9}{5}\right)$.

Find the slope of this line: $m = \frac{\frac{9}{5} - \frac{7}{5}}{\frac{6}{5} - \frac{7}{5}} = \frac{2/5}{-1/5} = -2$.

The equation for this line, using point-slope form with the point $\left(\frac{7}{5}, \frac{7}{5}\right)$ and slope $-2$, is:

$ \begin{aligned} &y - \frac{7}{5} = -2 \left( x - \frac{7}{5} \right) \\ &\Rightarrow 5y - 7 = -10x + 14 \\ &\Rightarrow 10x + 5y - 21 = 0 \end{aligned} $

So, the equation of the third side is $10x + 5y - 21 = 0$.

$ \begin{aligned} & \text { } \begin{aligned} \frac{\partial}{\partial x}\left(a x^2+2 h x y-2 a y^2\right. & +3 x+15 y-9) =2 a x+2 h y+3 \\ \frac{\partial}{\partial y}\left(a x^2+2 h x y-2 a y^2\right. & +3 x+15 y-9) = 2 h x-4 a y+15 \end{aligned} \end{aligned} $

$\Rightarrow$ At point $(1,1)$

$ \begin{aligned} & 2 a+2 h+3=0 \\ & 2 h-4 a+15=0 \end{aligned} $

Solving above equation, we get

$ \begin{aligned} & a=2, y=\frac{-7}{2} \\ \Rightarrow \quad & a h=2 \times \frac{-7}{2}=-7 \end{aligned} $

$ \text { Let } R \text { be }(h, k), A(2,3) $

$ P(h, 0), Q(0, k) $

Equation of line $\frac{x}{h}+\frac{y}{k}=1$

This is passes through $(2,3)$

$ \begin{aligned} \frac{2}{h}+\frac{3}{k} & =1 \\ 3 h+2 k-h k & =0 \end{aligned} $

$\therefore$ Required locus $3 x+2 y=x y$

Lines $x+2 a y+a=0$

$ \begin{aligned} & x+3 b y+b=0 \\ & x+4 c y+c=0 \end{aligned} $

These lines are concurrent

$ \begin{aligned} & \left|\begin{array}{lll} 1 & 2 a & a \\ 1 & 3 b & b \\ 1 & 4 c & c \end{array}\right|=0 \\ & \Rightarrow(3 b c-4 b c)-2 a(c-b)+a(4 c-3 b)=0 \\ & \Rightarrow-b c-2 a c+2 a b+4 a c-3 a b=0 \\ & \Rightarrow-b c+2 a c-a b=0 \\ & \Rightarrow 2 a c=a b+b c \end{aligned} $

$a, b, c$ are in HP

$ \begin{aligned} &\text { Foot of perpendicular from origin to }\\ &x-2 y+3=0 \end{aligned} $

$ \begin{gathered} \frac{x-0}{1}=\frac{y-0}{-2}=\frac{-3}{5} \\ x=\frac{-3}{5}, y=\frac{6}{5} \\ A \text { on }=\sqrt{\left(\frac{12}{5}\right)^2+\left(\frac{6}{5}\right)^2}=\frac{6 \sqrt{4+1}}{5}=\frac{6 \sqrt{5}}{5} \end{gathered} $

$x^2+x y-2 y^2=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ 3 y^2-5 x y-2 x^2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eq. (i), $x^2+2 x y-x y-2 y^2=0$

$ \begin{aligned} & (x-y)(x+2 y)=0 \\ & l_1=x-y=0 \\ & l_2=x+2 y=0 \end{aligned} $

From Eq. (ii), $3 y^2-5 x y-2 x^2=0$

$ \begin{aligned} \Rightarrow \quad & 3 y^2-6 x y+x y-2 x^2=0 \\ \Rightarrow \quad & (3 y+x)(y-2 x)=0 \\ & l_3 \equiv x+3 y=0 \\ & l_4 \equiv 2 x-y=0 \\ & l_2 \perp l_4 \end{aligned} $

$\therefore$ Non perpendicular lines

$ \begin{aligned} \Rightarrow \quad & (x-y)(x+3 y)=0 \\ \Rightarrow \quad & x^2+2 x y-3 y^2=a x^2+2 h x y+b y^2 \\ & a=1, h=1 \text { and } b=-3 \\ & a+2 h+b=1+2-3=0 \end{aligned} $

$ \begin{aligned} &\text { Given lines, }\\ &x+2 y+\lambda=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ 2 x^2-2 x y+3 y^2+2 x-y-1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) $\frac{x+2 y}{-\lambda}=1$

Now, homogenizing the Eq. (ii)

$ \begin{aligned} & \Rightarrow \quad 2 x^2-2 x y+3 y^2+(2 x-y)\left(\frac{x+2 y}{-\lambda}\right) \\ & \quad-\left(\frac{x+2 y}{-\lambda}\right)^2=6 \\ & \Rightarrow \quad\left(2 \lambda^2-2 \lambda-1\right) x^2+\left(-2 \lambda^2-3 \lambda-4\right) x y +\left(3 \lambda^2+2 \lambda-4\right) y^2=0 \end{aligned} $

For lines to be $\perp^{\prime} r$

$ \begin{aligned} \Rightarrow \quad & 2 \lambda^2-2 \lambda-1+3 \lambda^2+2 \lambda-4=0 \\ & \lambda= \pm 1 \Rightarrow \lambda=1 \end{aligned} $

Let $P(h, k)$

Given, $\left|\frac{2 h-3 k+1}{\sqrt{4+9}}\right|=2$

$ 2 h-3 k+1=2 \sqrt{13} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $\sqrt{(h-5)^2+(k-6)^2}=\sqrt{13}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii), we get

$ \left.2 h-3 k+1=2 \sqrt{\left[(h-15)^2+(k-6)^2\right.}\right] $

On squaring both sides, we get

$ \begin{aligned} & 4 h^2+9 k^2+1-12 h k-6 k \\ + & 4 h=4\left[h^2+25-10 h+k^2+36-12 k\right] \\ \Rightarrow & 5 k^2-12 h k+44 h+42 k-243=0 \\ \Rightarrow & 12 h k-5 k^2-44 h-42 k+243=0 \end{aligned} $

Taking locus of $P(h, k)$, we get

$ 12 x y-5 y^2-44 x-42 y+243=0 $

We have, $3 x^2+4 y^2-x y+k=0$ is transformed equation after shifting

$ \begin{aligned} &\text { origin to the point }(\alpha, \beta) \text { replacing }\\ &\begin{aligned} & \begin{aligned} & x \rightarrow x-\alpha \text { and } y \rightarrow y-\beta \\ & \therefore 3(x-\alpha)^2+4(y-\beta)^2-(x-\alpha)(y-\beta)+k=0 \\ & \Rightarrow 3\left(x^2+\alpha^2-2 \alpha x\right)+4\left(y^2+\beta^2-2 \beta y\right) \\ & \quad-x y+\alpha y+\beta x-\alpha \beta+k=0 \\ & \Rightarrow 3 x^2+4 y^2-x y+x(-6 \alpha+\beta) +y(-8 \beta+\alpha) \\ & \quad+3 \alpha^2+4 \beta^2-\alpha \beta+k=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} \end{aligned} \end{aligned} $

Comparing Eq. (i) with given equation,

$ \begin{aligned} & -6 \alpha+\beta=-5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \alpha-8 \beta=-7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & 3 \alpha^2+4 \beta^2-\alpha \beta+k=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

On solving Eqs. (ii) and (iii), we get $\alpha=\beta=1$

By Eqs. (ii)

$ \begin{array}{ll} & 3+4-1+k=2 \\ \Rightarrow & k=-4 \\ \therefore & \alpha+\beta-k=1+1+4=6 . \end{array} $

Let the $L$ makes and angle $\theta$ from positive direction of $X$-axis.

$\therefore$ The equation of $L$ is

$ \frac{x-1}{\cos \theta}=\frac{y-5}{\sin \theta}=r \text { and }-r $

Let $A(r \cos \theta+1, r \sin \theta+5)$ and $B(-r \cos \theta+1,-r \sin \theta+5)$

be two points which lies in the given two lines $5 x-y-4=0$ and $3 x+4 y-4=0$

because $L$ bisects these lines.

$ \begin{aligned} & 5(r \cos \theta+1)-r \sin \theta-5-4=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & 3(-r \cos \theta+1)+4(-r \sin \theta+5)-4=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & r(5 \cos \theta-\sin \theta)=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & r(-3 \cos \theta-4 \sin \theta)=-19 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

From Eqs. (iii) and (iv)

$ \begin{aligned} & -95 \cos \theta+19 \sin \theta=-12 \cos \theta-16 \sin \theta \\ & \Rightarrow \quad 35 \sin \theta=83 \cos \theta \\ & \therefore \quad \tan \theta=\frac{83}{35}=m \end{aligned} $

$\therefore$ Equation of $L$ is:

$ \begin{aligned} & y-5=\frac{83}{35}(x-1) \\ \Rightarrow \quad & 83 x-35 y+92=0 \end{aligned} $