Straight Lines and Pair of Straight Lines

$ \text {Orthocenter }(O):(0,0) \text {. Since it's an equilateral triangle, this is also the Centroid. } $

• Side BC Line : $x+2 \sqrt{2} y=4$.

• Vertex A : $(\alpha, \beta)$.

1. Find the distance from the Centroid to side BC

The side $B C$ lies on the line $x+2 \sqrt{2} y-4=0$. The distance from the origin $G(0,0)$ to this line is the length of the "lower" part of the median ( $r$ ).

Using the point-to-line distance formula :

$ r=\frac{|0+2 \sqrt{2}(0)-4|}{\sqrt{1^2+(2 \sqrt{2})^2}}=\frac{|-4|}{\sqrt{1+8}}=\frac{4}{3} $

2. Find the coordinates of Vertex $\mathbf{A}(\alpha, \beta)$

The centroid $G$ divides the segment $A D$ (where $D$ is the foot of the altitude on $B C$ ) in the ratio $2: 1$. Therefore, the distance from $A$ to $G$ is $2 r$.

$ A G=2 \times \frac{4}{3}=\frac{8}{3} $

The line $A G$ is perpendicular to side $B C$. The slope of $B C$ is $m=-\frac{1}{2 \sqrt{2}}$, so the slope of the line $A G$ (the altitude) is the negative reciprocal:

$ m_{\perp}=2 \sqrt{2} $

Since $A$ lies on the line passing through the origin with slope $2 \sqrt{2}$, its coordinates can be written as:

$ \alpha=k, \quad \beta=2 \sqrt{2} k $

Because $A$ and the side $B C$ are on opposite sides of the origin (the centroid), and the constant term in our line equation is negative, $A$ must satisfy the condition that its position yields an opposite sign. Substituting these into the distance formula $A G=\frac{8}{3}$ :

$ \sqrt{k^2+(2 \sqrt{2} k)^2}=\frac{8}{3} \Longrightarrow \sqrt{9 k^2}=\frac{8}{3} \Longrightarrow 3|k|=\frac{8}{3} \Longrightarrow|k|=\frac{8}{9} $

3. Calculate $|\alpha+\sqrt{2} \beta|$

Substitute $\alpha=k$ and $\beta=2 \sqrt{2} k$ :

$ |\alpha+\sqrt{2} \beta|=|k+\sqrt{2}(2 \sqrt{2} k)| $

$ =|k+4 k|=|5 k| $

$ =5 \times \frac{8}{9}=\frac{40}{9} \approx 4.44 $

greatest integer less than or equal to $|\alpha+\sqrt{2} \beta|=[4.44]=4$.

Let any point $Q(t, 6-t)$ on line $x+y=6$ such that.

distance $P Q=\sqrt{\frac{2}{3}} \Rightarrow P Q^2=\frac{2}{3}$.

using distance formula for points $P(2,3)$ and $Q(t, 6-t)$.

$ \begin{aligned} & (2-t)^2+(3-6+t)^2=\frac{2}{3} \\ & (2-t)^2+(t-3)^2=\frac{2}{3} \\ & 4+t^2-4 t+t^2+9-6 t=\frac{2}{3} \\ & 2 t^2-10 t+13=\frac{2}{3} \\ & 6 t^2-30 t+39-2=0 \\ & 6 t^2-30 t+37=0 \end{aligned} $

let $t_1 \& t_2$ are roots of this equation.

$ t_1+t_2=\frac{30}{6}=5 \text { and } t_1 t_2=\frac{37}{6} . $

So, there are two point $Q_1\left(t_1, 6-t_1\right) \& Q_2\left(t_2, 6-t_2\right)$.

It is given that angle made with positive x -axis by straight lines from $P(2,3)$ to $Q_1 \& Q_2$ is $\theta_1 \& \theta_2$ respectively.

This means $\tan \theta_1=$ slope of line $P Q_1$.

$ \tan \theta_1=\frac{3-\left(6-t_1\right)}{2-t_1} \Rightarrow \theta_1=\frac{t_1-3}{2-t_1} $

similarly, $\tan \theta_2=$ slope of line $P Q_2$

$ \tan \theta_2=\frac{t_2-3}{2-t_2} \Rightarrow \theta_2=\tan ^{-1}\left(\frac{t_2-3}{2-t_2}\right) $

adding equation (1) \& (2)

$ \begin{aligned} & \theta_1+\theta_2=\tan ^{-1}\left(\frac{t_1-3}{2-t_1}\right)+\tan ^{-1}\left(\frac{t_2-3}{2-t_2}\right) \\ & =\tan ^{-1}\left(\frac{\frac{t_1-3}{2-t_1}+\frac{t_2-3}{2-t_2}}{1-\left(\frac{t_1-3}{2-t_1}\right)\left(\frac{t_2-3}{2-t_2}\right)}\right), \text { use } \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ & =\tan ^{-1} \frac{2 t_1-6-t_1 t_2+3 t_1+2 t_2-6-t_1 t_2+3 t_1}{4-2\left(t_1+t_2\right)+t_1 t_2-\left(9-3\left(t_1+t_2\right)+t_1 t_2\right)} \\ & \theta_1+\theta_2=\tan ^{-1} \frac{5\left(t_1+t_2\right)-2 t_1 t_2-12}{\left(t_1+t_2\right)-5} \end{aligned} $

substitute value of $t_1+t_2=5 \& t_1 t_2=\frac{37}{6}$

$ \begin{aligned} & \theta_1+\theta_2=\tan ^{-1} \frac{5 \times 5-2 \times \frac{37}{6}-12}{5-5} \\ & \theta_1+\theta_2=\tan ^{-1}(\infty)=\frac{\pi}{2} \end{aligned} $

solution : $A(1,0), B(2,-1)$ and $C\left(\frac{7}{3}, \frac{4}{3}\right)$

equation of line AB using two point form

$ \begin{aligned} & y-0=\frac{0-(-1)}{1-2}(x-1) \\ & y=-x+1 \\ & L_1: x+y-1=0 \end{aligned} $

similarly equation of line $B C$.

$ \begin{aligned} & y+1=\frac{\frac{4}{3}-(-1)}{\frac{7}{3}-2}(x-2) \\ & y+1=\frac{7 / 3}{1 / 3}(x-2) \\ & y=7 x-14-1 \\ & L_2: 7 x-y-15=0 \end{aligned} $

from graph ( 2,0 ) lies in the region of $\angle A B C$.

calculating $L_1(2,0) \cdot L_2(2,0)$

$ (2+0-1) \cdot(7 \times 2-0-15)=1 \cdot(-1)=-1 $

equation of angular bisector of $\angle A B C$ is given as

$ \begin{aligned} & \frac{x+y-1}{\sqrt{1^2+1^2}}= \pm \frac{7 x-y-15}{\sqrt{7^2+(-1)^2}} \\ & \frac{x+y-1}{\sqrt{2}}= \pm \frac{7 x-y-15}{\sqrt{50}} \\ & 5 x+5 y-5= \pm(7 x-y-15) \end{aligned} $

since $L_1(2,0) \cdot L_2(2,0)$ is negative so equation of $\angle A B C$ bisector in the region $(2,0)$ can be

find using negative sign from equation (1)

$ \begin{aligned} & 5 x+5 y-5=-(7 x-y-15) \\ & 5 x+5 y-5+7 x-y-15=0 \\ & 12 x+4 y-20=0 \end{aligned} $

$3 x+y=5$ comparing with bisector $\alpha x+\beta y=5$.

$\alpha=3$ and $\beta=1$

$\alpha^2+\beta^2=9+1=10$.

Let $A=(1,2)$ and $C=(-3,-6)$ be diagonally opposite vertices of a rhombus.

In a rhombus, the diagonals bisect each other at a common point M..

Calculation of Midpoint $M$ :

$ \begin{aligned} & M=\left(\frac{1+(-3)}{2}, \frac{2+(-6)}{2}\right) \\ & M=\left(\frac{-2}{2}, \frac{-4}{2}\right) \\ & M=(-1,-2) \end{aligned} $

Since $M$ is also the midpoint of the other diagonal $B D$ where $B=(\alpha, \beta)$ and $D=(\gamma, \delta)$ :

$ \begin{aligned} & \frac{\alpha+\gamma}{2}=-1 \Longrightarrow \alpha+\gamma=-2 \\ & \frac{\beta+\delta}{2}=-2 \Longrightarrow \beta+\delta=-4 \end{aligned} $

$ \begin{aligned} & |\alpha+\beta+\gamma+\delta|=|(\alpha+\gamma)+(\beta+\delta)| \\ & |\alpha+\beta+\gamma+\delta|=|(-2)+(-4)| \\ & |\alpha+\beta+\gamma+\delta|=|-6| \\ & |\alpha+\beta+\gamma+\delta|=6 \end{aligned} $

Option (B) is correct.

For $L$ to bisect area, line must pass through the centre of rectangle $\left(\frac{3}{2}, 2\right)$

$ \begin{aligned} & \Rightarrow \quad L \equiv(y-2)=\frac{1}{3}\left(x-\frac{3}{2}\right) \\ & \Rightarrow \quad 3 y-6=x-\frac{3}{2} \Rightarrow 6 y-12=2 x-3 \\ & 6 y-2 x-9=0 \end{aligned} $

The distance from $L$ of $\left(\frac{1}{2},-5\right)$ is

$ \begin{aligned} & \left|\frac{6(-5)-2\left(\frac{1}{2}\right)-9}{\sqrt{40}}\right|=\sqrt{40} \\ & =2 \sqrt{10} \end{aligned} $

$ \begin{aligned} &\text { Solution of statement-I }\\ &m_{A O} m_{B C}=-1 \end{aligned} $

$ \begin{aligned} & \Rightarrow 2 h-3 k=13 \ldots \ldots \\ & \& m_{A B} m_{O C}=-1 \\ & \Rightarrow 4 k=7 h \ldots \ldots \ldots \ldots . \end{aligned} $

⇒ third vertex is $(-4,-7)$

Solution of statement -2

$ \begin{aligned} & 2 a, b, c \rightarrow A \cdot P \\ & b=\frac{2 a+c}{2} \Rightarrow 2 a-2 b+c=0 \end{aligned} $

∴ lines $a x+b y+c=0$ are concurrent then

$ \begin{aligned} & \frac{x}{2}=\frac{y}{-2}=\frac{1}{1} \\ & x=2 \text { and } y=-2 \end{aligned} $

$\therefore \mathrm{Pt}$. Of concurrency is $(2,-2)$

∴ Statement 2 is correct.

$ \text { Let } a \text { be the side of } \triangle A B C $

$ \begin{aligned} & \sin \theta=\frac{3}{a} \\ & \sin \left(60^{\circ}+\theta\right)=\frac{9}{a} \\ & \Rightarrow \frac{\sqrt{3}}{2} \times \sqrt{1-\frac{9}{a^2}}+\frac{3}{a} \times \frac{1}{2}=\frac{9}{a} \end{aligned} $

$ \begin{aligned} & \Rightarrow \sqrt{3}\left(\sqrt{a^2-9}\right)+3=18 \\ & \Rightarrow 3\left(a^2-9\right)=15^2 \quad \Rightarrow a^2-9=15 \times 5 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a^2=84 \\ & \text { Area }=\frac{\sqrt{3}}{4} a^2=21 \sqrt{3} \end{aligned} $

$\begin{aligned} &\begin{aligned} & \text { Slope of diagonal } \mathrm{OB}=\frac{\sqrt{3}+1}{1-\sqrt{3}} \\ & =\tan 105^{\circ} \\ & \therefore \alpha=60^{\circ} \\ & \therefore \mathrm{A}\left(\operatorname{acos} 60^{\circ}, \mathrm{asin} 60^{\circ}\right) \\ & \therefore \mathrm{A}\left(\frac{\mathrm{a}}{2}, \frac{\sqrt{3} \mathrm{a}}{2}\right) \end{aligned}\\ &\text { A Lies on other diagonal }\\ &\begin{aligned} & \therefore\left(\frac{\sqrt{3}-1}{2}\right) a-\left(\frac{\sqrt{3}+1}{2}\right) \cdot \sqrt{3} a+8 \sqrt{3}=0 \\ & a\left[\frac{\sqrt{3}-1-3-\sqrt{3}}{2}\right]=-8 \sqrt{3} \end{aligned} \end{aligned}$

$\begin{aligned} & \mathrm{a}=4 \sqrt{3} \\ & \therefore \mathrm{a}^2=48 \end{aligned}$

$\begin{aligned} & \mathrm{m}_{\mathrm{PR}}=2-\sqrt{3}=\tan 15^{\circ} \\ & \therefore \frac{\alpha}{2}=15^{\circ} \end{aligned}$

$\Rightarrow \alpha=30^{\circ}$

equation of PR:

$\begin{aligned} & y=\tan 15^{\circ}(x-a) \\ & y=(2-\sqrt{3})(x-a) \end{aligned}$

$\perp$ distance from origin $=\frac{1}{\sqrt{2}}$

$\begin{aligned} & \left|\frac{\sqrt{3} a-2 a}{\sqrt{4+3-4 \sqrt{3}+1}}\right|=\frac{1}{\sqrt{2}} \\ & \frac{|a|(2-\sqrt{3})}{2 \sqrt{(2-\sqrt{3})}}=\frac{1}{\sqrt{2}} \\ & |a|=\frac{\sqrt{2}}{\sqrt{2-\sqrt{3}}}=\sqrt{2}(\sqrt{2+\sqrt{3}}) \\ & a^2=2(2+\sqrt{3}) \\ & 3 a^2 \tan ^2 \alpha-2 \sqrt{3} \\ & 3 \times(4+2 \sqrt{3}) \cdot \frac{1}{3}-2 \sqrt{3}=4 \end{aligned}$

Solve line PQ & QR

$\begin{aligned} & \text { Point } \mathrm{Q}\left(\frac{1-\mathrm{c}}{\mathrm{~m}-1}, \frac{1-\mathrm{c}}{\mathrm{~m}-1}+1\right) \\ & \mathrm{m}_{2 \mathrm{H}}=\frac{\frac{1-\mathrm{c}}{\mathrm{~m}-1}+2}{\frac{1-\mathrm{c}}{\mathrm{~m}-1}-3}=\frac{1-\mathrm{c}+2 \mathrm{~m}-2}{1-\mathrm{c}-3 \mathrm{~m}+3}=-\frac{1}{4} \quad\text{..... (1)}\\ & \because \mathrm{~m}_{\mathrm{PH}}=\frac{5}{0} \rightarrow \infty \\ & \Rightarrow \text { Slope of line } \mathrm{QR}(\mathrm{~m})=0 \end{aligned}$

Put value of $m$ in equation (1)

$\frac{1-c-2}{1-c+3}=-\frac{1}{4} \Rightarrow c=0$

so $\mathrm{m}-\mathrm{c}=0$ Ans.

Equation of line $A B$ is $3 y-x=2$

And $A C$ is $x+y=2$

In line $A B$,

When $y=0, x=-2$

$\therefore \quad B(-2,0)$

In line $A C$,

When $y=0, x=2$

$\therefore \quad C(2 x, 0)$

Equation of altitude of $B C$,

$Y=x+2$

Similarly, equation of altitude of $A B$,

$y=-3 x+6$

$\therefore$ On solving, orthocentre $P(1,3)$

$\therefore \quad \operatorname{ar}(\triangle P B C)=6$

A is orthocentre of above $\Delta$.

$\begin{aligned} &O \text { is orthocentre of above } \Delta \text {. }\\ &O A=\sqrt{5} \end{aligned}$

$\begin{aligned} & x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5 \\ & (x+2 y-5)+\lambda(3 x+7 y-17)=0 \\ & L_1+\lambda L_2=0 \\ & \Rightarrow \quad P \text { is intersection of } L_1 \& L_2 \text { i.e. }(1,2) \\ & y-2=m(x-1) \\ & m x-y+2-m=0 \\ & \text { Distance from origin }=\left|\frac{2-m}{\sqrt{1+m^2}}\right|=\max \end{aligned}$

$\begin{aligned} & \text { For } m=-\frac{1}{2} \\ & \therefore L=y-2=\frac{-1}{2}(x-1) \\ & L: x+2 y-5=0 \\ & \text { Now, } d=\left|\frac{3+12-5}{\sqrt{5}}\right|=\left|\frac{10}{\sqrt{5}}\right| \\ & d^2=\frac{100}{5}=20 \end{aligned}$

$\begin{aligned} & \tan \theta=\left|\frac{-2-1}{1+(-2)(1)}\right|=\frac{3}{1} \\ & \sin \theta=\frac{3}{\sqrt{10}} \\ & A M=\frac{3}{\sqrt{10}} \times \frac{9}{\sqrt{2}}=\frac{27}{2 \sqrt{5}} \end{aligned}$

Dist. between $I_1 ~\& ~l_2, \quad A M=\left|\frac{\frac{P}{2}+6}{\sqrt{5}}\right|=\frac{27}{2 \sqrt{5}}$

$\begin{aligned} & \frac{P}{2}+6= \pm \frac{27}{2} \\ & \frac{P}{2}=\frac{27}{2}-6 \Rightarrow P=15, \text { As } P>0 \\ & B M=\sqrt{\frac{81}{2}-\left(\frac{27}{2 \sqrt{5}}\right)^2} \\ & =\sqrt{\frac{810-729}{20}}=\sqrt{\frac{81}{20}}=\frac{9}{2 \sqrt{5}} \\ & \text { Now, } \frac{A M}{B M}=\frac{\frac{27}{\frac{2 \sqrt{5}}{9}}}{\frac{2 \sqrt{5}}{2 \sqrt{2}}}=3 \end{aligned}$

$\begin{aligned} & L: x+b y+c=0 \\ & \because \frac{1}{2}\left|(-c) \cdot\left(\frac{-c}{b}\right)\right|=48 \\ & \therefore\left|\frac{c^2}{b}\right|=96\quad\text{..... (i)} \end{aligned}$

Slope of line $L=-\frac{1}{b}$

$\therefore$ Slope of line perpendicular to $L$ is $b$.

$\begin{aligned} & \therefore b=1 \\ & \therefore c^2=96 \\ & \therefore b^2+c^2=97 \end{aligned}$

Area of $\triangle \mathrm{AOB}=\frac{1}{2}$

Area of $\triangle \mathrm{AMN}=\frac{4}{9} \times \frac{1}{2}=\frac{2}{9}$

Equation of AB is $\mathrm{x}+\mathrm{y}=1$

$\begin{aligned} & \mathrm{OA}=1, \mathrm{AM}=\sec \left(45^{\circ}-\theta\right) \\ & \mathrm{AN}=\sec \left(45^{\circ}-\theta\right) \cos \theta \\ & \mathrm{MN}=\sec \left(45^{\circ}-\theta\right) \sin \theta \end{aligned}$

$\begin{aligned} & \operatorname{Ar}(\triangle \mathrm{AMN})=\frac{1}{2} \times \sec ^2\left(45^{\circ}-\theta\right) \sin \theta \cdot \cos \theta=\frac{2}{9} \\ & \Rightarrow \tan \theta=2, \frac{1}{2} \\ & \tan \theta=2 \text { is rejected } \\ & \frac{\mathrm{AN}}{\mathrm{NB}}=\frac{\lambda}{1}=\cot \theta=2 \end{aligned}$

$\begin{aligned} \therefore \text { centroid of } \triangle \mathrm{ABC} & =\left(\frac{9+3+5}{3}, \frac{11+4+13}{3}\right) \\ & =\left(\frac{17}{3}, \frac{28}{3}\right) \end{aligned}$

Let image of centroid with respect to line mirror is (h, k)

$\begin{aligned} &\begin{aligned} & \therefore\left(\frac{\mathrm{k}-\frac{28}{3}}{\mathrm{~h}-\frac{17}{3}}\right)\left(-\frac{1}{2}\right)=-1 \\ & \& 3\left(\frac{\mathrm{~h}+\frac{17}{3}}{2}\right)+6\left(\frac{\mathrm{k}+\frac{28}{3}}{2}\right)=53 \end{aligned}\\ &\text { Solving (1) \& (2) we get } \mathrm{h}=3, \mathrm{k}=4\\ &\therefore \mathrm{h}^2+\mathrm{k}^2+\mathrm{hk}=37 \end{aligned}$

$\begin{aligned} & \tan \theta=\frac{m-\frac{1}{2}}{1+\frac{1}{2} \cdot m}=\frac{-1-m}{1-m}=\frac{m+1}{m-1} \\ & \frac{2 m-1}{2+m}=\frac{m+1}{m-1} \\ & 2 m^2-3 m+1=m^2+3 m+2 \end{aligned}$

$\begin{aligned} & m^2-6 m-1=0 \\ & \text { sum of root }=6 \\ & \text { sum is } 6 \end{aligned}$

$\begin{aligned} & x^2+y^2-8 x=0, \frac{x^2}{9}-\frac{y^2}{4}=1 \quad\text{.... (1)}\\ & 4 x^2-9 y^2=36 \quad\text{.... (2)}\\ & \text { Solve }(1) \&(2) \\ & 4 x^2-9\left(8 x-x^2\right)=36 \\ & 13 x^2-72 x-36=0 \\ & (13 x+6)(x=6)=0 \\ & x=\frac{-6}{13}, x=6 \\ & x=\frac{-6}{13}(\text { rejected }) \\ & y \rightarrow \text { Imaginary } \\ & n=6, \frac{36}{9}-\frac{y^2}{4}=1 \\ & y^2=12, y=I \sqrt{12} \\ & A(6, \sqrt{12}), B(6,-\sqrt{12}) \\ & p\left(\alpha, \frac{2 \alpha+4}{3}\right) P \text { lies on } \end{aligned}$

$\begin{aligned} & \text { centroid }(\mathrm{h}, \mathrm{k}) \quad 2x-3y+y=0\\ & \mathrm{h}=\frac{12+\alpha}{3}, \alpha=3 \mathrm{~h}-12 \\ & \mathrm{k}=\frac{\frac{2 \alpha-3 y}{3}}{3} \Rightarrow 2 \alpha+4=9 \mathrm{y} \\ & \alpha=\frac{9 \mathrm{k}-4}{2} \\ & 6 \mathrm{~h}-2 \mathrm{y}=9 \mathrm{k}-4 \\ & 6 \mathrm{x}-9 \mathrm{y}=20 \end{aligned}$

Point of intersection of $x=\frac{11}{2}$ with $L_1 \& L_3$ gives,

$\alpha_{\min }=\frac{11}{2}$

and $\alpha_{\max }=6$

$\therefore \alpha_{\min } \cdot \alpha_{\max }=\frac{11}{2} \times 6=33$

$\because \mathrm{PM}=\mathrm{QM}$

So, $M\left(\frac{\frac{57}{13}+1}{2}, \frac{\frac{-40}{13}+2}{2}\right)$

$=\left(\frac{35}{13}, \frac{-7}{13}\right)$

$\because \mathrm{M}$ lies on the time

$\begin{aligned} & 2 x-3 y+\lambda=0 \\ & 2\left(\frac{35}{13}\right)-3\left(\frac{-7}{13}\right)+\lambda=0 \\ & \lambda=-\frac{70}{13}+\frac{21}{13} \\ & =\frac{-91}{13}=-7 \end{aligned}$

$\begin{aligned} & \left|\begin{array}{ccc} 3 & -4 & -\alpha \\ 8 & -11 & -33 \\ 2 & 3 & \lambda \end{array}\right|=0 \\ & \Rightarrow 3(-11 \lambda-99)+4(8 \lambda+66)-\alpha(-24+22)=0 \\ & \Rightarrow 33 \lambda-297+32 \lambda+264+24 \alpha-22 \alpha=0 \\ & \Rightarrow-\lambda+2 \alpha-33=0 \quad\text{.... (1)}\\ & \therefore \lambda=-7 \\ & -(-7)+2 \alpha-33=0 \\ & 2 \alpha=26 \\ & \alpha=13 \\ & \therefore|\alpha \lambda|=|13 \times(-7)| \\ & =91 \end{aligned}$

$\begin{aligned} & \mathrm{h}=\frac{3 \beta+\alpha}{3} \\ & \mathrm{k}=\frac{-4+\alpha+2}{3} \\ & \alpha=3 \mathrm{k}+2 \\ & 2 \beta=3 \mathrm{~h}-\mathrm{a}=3 \mathrm{~h}-3 \mathrm{k}-2 \\ & \text { so } \mathrm{AB}=8 \\ & (\alpha-\beta)^2+(\alpha+4)^2=64 \\ & \left(3 \mathrm{k}+2-\left(\frac{3 \mathrm{~h}-3 \mathrm{k}-2}{2}\right)\right)^2+(3 \mathrm{k}+2+4)^2=64 \\ & \frac{(9 \mathrm{k}-3 \mathrm{~h}+6)^2}{4}+(3 \mathrm{k}+6)^2=64 \\ & 9\left[(3 \mathrm{k}-\mathrm{h}+2)^2+4(\mathrm{k}+2)^2\right]=64 \times 4 \\ & 9\left(\mathrm{x}^2+13 \mathrm{y}^2-6 \mathrm{xy}-4 \mathrm{x}+28 \mathrm{y}\right)=76 \\ & \alpha-\beta-\gamma=13+6+4=23 \end{aligned}$

Let ' $G$ ' be the centroid of $\Delta$ formed by $(1,3)(3,1)$ \& $(2,4)$

$ \mathrm{G} \cong\left(2, \frac{8}{3}\right) $

Image of G w.r.t. $x+2 y-2=0$

$ \begin{aligned} & \frac{\alpha-2}{1}=\frac{\beta-\frac{8}{3}}{2}=-2 \frac{\left(2+\frac{16}{3}-2\right)}{1+4} \\ & =\frac{-2}{5}\left(\frac{16}{3}\right) \\ & \Rightarrow \alpha=\frac{-32}{15}+2=\frac{-2}{15}, \beta=\frac{-32 \times 2}{15}+\frac{8}{3}=\frac{-24}{15} \\ & 15(\alpha-\beta)=-2+24=22 \end{aligned} $

$\left.\begin{array}{c}4 x-3 y=12 \alpha \\ 4 x+3 y=\frac{12}{\alpha}\end{array}\right\} \rightarrow 16 x^2-9 y^2=144$

$\begin{aligned} & \frac{x^2}{9}-\frac{y^2}{16}=1 \rightarrow \text { Curve :S } \\ & T: y=m x \pm \sqrt{9 m^2-16} \\ & m=\frac{4 \sqrt{2}}{3} \\ & y=\frac{4 \sqrt{2} x}{3} \pm \sqrt{32-16} \\ & 3 y=4 \sqrt{2} x \pm 12 \\ & \text { as } q>0 \\ & 3 y=4 \sqrt{2} x+12\end{aligned}$

$ \mathrm{p}=-\frac{3}{\sqrt{2}} \quad \& \quad \mathrm{q}=4 $

$ \mathrm{pq}=-6 \sqrt{2} $

Equation of $A B: y-2=-1(x)$

Equation of $B C: y-2=1(x)$

Equation of $A C$ : $y-0=0(x-2)$

Let the variable point be $(x, y)$.

$ \begin{gathered} a=\frac{|x+y-2|}{\sqrt{2}}, \\ b=\frac{|x-y+2|}{\sqrt{2}} \\ c=|y| \\ 2 b=a+c \\ 2\left|\frac{x-y+2}{\sqrt{2}}\right|=\left|\frac{x+y-2}{\sqrt{2}}\right|+|y| \end{gathered} $

Hence lows is

$ |\sqrt{2} y|=2|x-y+2|-|x+y-2| $

$ \begin{aligned} & \text { } 4 a^2+9 b^2-12 a b-c^2=0 \\ & \Rightarrow(2 a-3 b)^2-c^2=0 \\ & \Rightarrow(2 a-3 b-c)(2 a-3 b+c)=0 \\ & \Rightarrow 2 a-3 b-c=0 \text { or } 2 a-3 b+c=0 \end{aligned} $

$\Rightarrow a x+b y+c$ is passing through $(-2, 3)$ and also passing through ( $2,-3$ )

Let slope of line $A B$ be -1 and slope of $A D$ be 7

$ \begin{aligned} & \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{7-(-1)}{1+7 \times-1}\right| \\ & \tan \theta=\frac{8}{6}=\frac{4}{3} \end{aligned} $

Let slope of $A C$ be $m$.

$ \begin{aligned} & \Rightarrow\left|\frac{-1-m}{1-m}\right|=\left|\frac{m-7}{1+7 m}\right| \\ & \Rightarrow \frac{-1-m}{1-m}= \pm\left(\frac{m-7}{1+7 m}\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow(-1-m)(1+7 m)=(1-m)(m-7) \\ & \Rightarrow-1-7 m-m-7 m^2=m-7-m^2+7 m \\ & \Rightarrow-7 m^2-8 m-1=-m^2+8 m-7 \\ & \Rightarrow 6 m^2+16 m-6=0 \\ & \Rightarrow 3 m^2+8 m-3=0 \\ & \Rightarrow 3 m^2+9 m-m-3=0 \\ & \Rightarrow 3 m(m+3)-1(m+3)=0 \\ & \Rightarrow m=-3 \text { or } m=\frac{1}{3} \end{aligned} $

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

$ \begin{aligned} & y-3=-3 x+3 \\ & \Rightarrow 3 x+y-6=0 \end{aligned} $

Solving $3 x+y-6=0$ and $15 x-5 y=6$

we get $\alpha$ and $\beta$

$ \begin{aligned} & 3 x+y=6 \\ & \Rightarrow 3 x-y=\frac{6}{5} \Rightarrow 2 y=6-\frac{6}{5} \\ & \Rightarrow y=\frac{12}{5}=\beta \Rightarrow 3 x=\frac{6-12}{5} \\ & \Rightarrow 3 x=\frac{18}{5} \Rightarrow x=\frac{6}{5}=\alpha \\ & \Rightarrow \alpha+\beta=\frac{18}{5} \end{aligned} $

The first given equation is $3x^2 + 2hxy - 3y^2 = 0$. This equation represents two straight lines passing through the origin because there are no constant terms.

This equation can be written as the product of two linear factors: $3x^2 + 2hxy - 3y^2 = (3x - y)(x + 3y) = 0$, which means $h=4$.

The second equation is $3x^2 + 2hxy - 3y^2 + 2x - 4y + c = 0$. This also represents two straight lines, but they are not passing through the origin.

Since both equations together represent the four sides of a square, the sides must be parallel and equal in pairs. Therefore, the second equation must be the same as the first one but shifted by some constants.

The second equation can be written as $(3x - y + c_1)(x + 3y + c_2) = 0$ for some $c_1$ and $c_2$.

Now, expanding: $(3x - y + c_1)(x + 3y + c_2)$ gives $3x^2 + 2xy - 3y^2 + (3c_2 + c_1)x + (3c_1 - c_2)y + (c_1c_2)$.

Comparing this with $3x^2 + 2hxy - 3y^2 + 2x - 4y + c = 0$, we match coefficients:

By comparing the $x$ terms: $3c_2 + c_1 = 2$.

By comparing the $y$ terms: $3c_1 - c_2 = -4$.

Solving these two equations:
$3c_2 + c_1 = 2$ … (i)
$3c_1 - c_2 = -4$ … (ii)

Multiply (i) by 3: $9c_2 + 3c_1 = 6$.
Add (ii): $3c_1 - c_2 = -4$
$9c_2 + 3c_1 + 3c_1 - c_2 = 6 -4$
$8c_2 + 6c_1 = 2$

Insert the solution from the original explanation: $c_1 = -1$ and $c_2 = 1$.

The constant term is $c = c_1c_2 = -1 \times 1 = -1$.

From earlier, we found $h = 4$, so $\frac{h}{c} = \frac{4}{-1} = -4$.

$ \begin{aligned} & \text { }(2,3) \xrightarrow[\text { shifted to }(3,2)]{\text { Origin }}(a, b) \\ & \Rightarrow \quad 3+a=2 \text { and } b+2=3 \\ & \Rightarrow \quad a=-1 \text { and } b=1 \end{aligned} $

$ \begin{aligned} &\begin{array}{lll} \hline & -1 & 1 \\ \hline c & \cos \theta & \sin \theta \\ \hline d & -\sin \theta & \cos \theta \\ \hline \end{array}\\ &\begin{gathered} c=-\cos \theta+\sin \theta \\ d=+\sin \theta+\cos \theta \\ -c+d=2 \cos \theta=2 \times \frac{1}{\sqrt{2}}=\sqrt{2} \end{gathered} \end{aligned} $

Solving the lines equation, we get point $A, B$ and $C$.

$ \begin{aligned} & \left.\begin{array}{l} x+y+4=0 \\ x-2 y-4=0 \end{array}\right\} \Rightarrow \begin{array}{l} C_x=\frac{-4}{3} \\ C_y=\frac{-8}{3} \end{array} \\ & \left.\begin{array}{l} x+y+4=0 \\ 3 x+4 y-2=0 \end{array}\right\} \begin{array}{l} B_x=-18 \\ B_y=14 \end{array} \\ & \left.\begin{array}{l} 3 x+4 y-2=0 \\ x-2 y-4=0 \end{array}\right\} \begin{array}{l} A_x=2 \\ A_y=-1 \end{array} \end{aligned} $

$A B \neq A C \neq B C \Rightarrow \triangle A B C$ is a scalene triangle.

$ \begin{aligned} &\text { Let equation of line } L=0 \text { be }\\ &y=m x+c \end{aligned} $

$ \begin{aligned} & 4=12 m+c \\ & \Rightarrow y=m x+4-12 m \end{aligned} $

⇒ Length of $X$-intercept of $L=\left|\frac{12 m-4}{m}\right|$

Length of $Y$-intercept of $L=|12 m-4|$

$ \begin{aligned} & \frac{1}{2} \times \frac{(12 m-4)^2}{|m|}=12 \\ & \Rightarrow \quad 8(3 m-1)^2=12 m(m>0) \\ & \Rightarrow \quad 2(3 m-1)^2=3 m \\ & \Rightarrow \quad 18 m^2-12 m+2=3 m \\ & \Rightarrow \quad 18 m^2-15 m+2=0 \\ & \Rightarrow \quad 18 m^2-3 m-12 m+2=0 \\ & \Rightarrow \quad 3 m(6 m-1)-2(6 m-1)=0 \\ & \quad m=\frac{1}{6} \text { or } m=\frac{2}{3} . \end{aligned} $

$ \begin{aligned} &\text { As intercept of } L \text { on } X \text {-axis is negative }\\ &\begin{aligned} \Rightarrow & m=\frac{1}{6}\left(m<\frac{1}{3}\right) \\ \Rightarrow & P=\frac{12 m-4}{m} \times\{(4-12 m)\}^2 \\ \Rightarrow & P=(12 m-4)^3 \times \frac{1}{m} \\ & =\left(12 \times \frac{1}{6}-4\right)^3 \times 6 \\ & =-8 \times 6=-48 \end{aligned} \end{aligned} $

Quadrilateral is a parallelogram.

∴ Area of parallelogram

$ \begin{aligned} & =\frac{\left|c_1-c_2\right| \times\left|d_1-d_2\right|}{\left|a_1 b_2-a_2 b_1\right|} \\ & =\frac{\left|\frac{9}{2}-3\right| \times\left|3-\frac{11}{3}\right|}{|1 \times-2-1 \times 2|} \\ & =\frac{3 \times 2}{2 \times 3 \times 4}=\frac{1}{4} \end{aligned} $

Let

$ \begin{aligned} & S=2 x^2+5 x y-3 y^2+2 g x+2 f y+C \\ & \Rightarrow \quad \frac{\partial S}{\partial x}=4 x+5 y+2 g \\ & \Rightarrow \quad \frac{\partial S}{\partial y}=5 x-6 y+2 f \end{aligned} $

$[\because(-1,-1)$ is the point in intersection]

$ \begin{array}{ll} \Rightarrow & 4 \times(-1)-5 \times 1+2 g=0 \\ \Rightarrow & 2 g=9 \\ \text { and } & 5 \times(-1)-6 \times(-1)+2 f=0 \\ & 2 f=-1 \\ & g+f=4 \end{array} $

Also, $(-1,-1)$ will satisfy

$ \begin{aligned} & 2 x^2+5 x y-3 y^2+2 g x+2 f y+c=0 \\ & \Rightarrow \quad 2+5-3-2 g-2 f+c=0 \\ & \Rightarrow \quad 4-8+c=0 \\ & \,\,\,\,\,\,\,\,\, \quad c=4 \\ & \Rightarrow \quad g+f=c \end{aligned} $

$ \begin{aligned} &h=\frac{3 a}{5}, k=\frac{2 b}{5}\\ &\text { ∴ Equation of } A B\\ &\frac{3}{a}+\frac{2}{b}-1=0 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \frac{3}{\frac{5 h}{3}}+\frac{2}{5 k}=1 \quad\left[\because h=\frac{3 a}{5}, k=\frac{2 b}{5}\right] \\ & \frac{9}{h}+\frac{4}{k}=5 \end{aligned}\\ &\text { Locus, }\\ &\begin{aligned} & \frac{9}{x}+\frac{4}{y}=5 \\ & 4 x+9 y=5 x y \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned} & X= x-h, Y=y-k \\ & X= x+1, Y=y-2 \\ & x=X-1, y=Y+2 \\ & \Rightarrow \quad 2(X-1)^2-(X-1)(Y+2)+(Y+2)^2-3(X-1)+4(Y+2) -5=0\end{aligned} \end{aligned} $

$ \begin{gathered} \Rightarrow \quad 2 X^2-4 X+2-X Y-2 X+Y+2 \\ +Y^2+4 y+4 \\ -3 X+3+4 Y+8-5=0 \end{gathered} $

$ \begin{array}{lr} 2 h=-1 & a=2 \\ 2 g=-4-2-3=-9 & b=1 \\ 2 f=1+4+4=9 & c=2+2+4+3+8-5 \\ & =14 \end{array} $

$ \begin{array}{ll} \therefore & 2(f+g+h)=-1 \\ \text { And } & c-5(a+b)=14-15=-1 \end{array} $

Using parametric form of line

$ \begin{aligned} & \frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}= \pm r \\ \Rightarrow \quad & \frac{x+2}{\cos 60^{\circ}}=\frac{y-4}{\sin 60^{\circ}}= \pm 6 \\ \Rightarrow & x= \pm 6 \cos 60^{\circ}-2, y= \pm 6 \sin 60^{\circ}+4 \\ & x= \pm 6\left(\frac{1}{2}\right)-2 y= \pm 3 \sqrt{3}+4 \\ = & \pm 3-2 \end{aligned} $

∵ Point lies in 3rd 24

$ \begin{aligned} & x=-5, y=-3 \sqrt{3}+4 \\ & p=-5, q=4-3 \sqrt{3} \end{aligned} $

$ \begin{aligned} & \therefore \sqrt{p^2+q^2-8 q} \\ & \quad=\sqrt{25+16+27-24 \sqrt{3}-32+24 \sqrt{3}} \\ & \quad=\sqrt{36}=6 \end{aligned} $

Foot of perpendicular of point $(2,-3)$ with respect to line $4 x-3 y+8=0$

$ \begin{aligned} & \frac{a-2}{4}=\frac{b+3}{-3}=\frac{-1(4(2)-3(-3)+8)}{4^2+3^2} \\ \Rightarrow & \frac{a-2}{4}=\frac{b+3}{-3}=-1 \\ \Rightarrow & x=-2, b=0 \\ \therefore & \quad a^3-b^3=k^3 \Rightarrow k=-2 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } Q(\alpha, \beta) \text { and } R(\gamma, \delta), P(1,2) \\ & \therefore \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{-2(1+2+1)}{1+1} \\ & \Rightarrow \frac{\alpha-1}{1}=\frac{\beta-2}{1}=-4 \Rightarrow \alpha=-3, \beta=-2 \end{aligned} \end{aligned} $

$ \begin{aligned} & \therefore Q=(-3,-2) \\ & \text { And } \frac{\gamma+3}{1}=\frac{\delta+2}{-1}=\frac{-2(-3+2-1)}{1+1} \\ & \Rightarrow \frac{\gamma+3}{1}=\frac{\delta+2}{-1}=2 \Rightarrow \gamma=-1, \delta=-4 \\ & R(-1,-4) \\ & \qquad M=\left(\frac{1-3}{2}, \frac{2-2}{2}\right)=(-1,0) \\ & \quad N=\left(\frac{-3-1}{2}, \frac{-2-4}{2}\right)=(-2,-3) \\ & \quad M N=\sqrt{1+9}=\sqrt{10} \end{aligned} $

$6 x^2+2 h x y+4 y^2=0$

Let slopes be $m_1$ of $m_2$

$ \begin{array}{ll} \therefore & \frac{m_1}{m_2}=\frac{2}{3} \\ \text { And } & m_1 m_2=\frac{a}{b}=\frac{6}{4} \\ & m_1 \cdot \frac{3 m_1}{2}=\frac{6}{4} \\ & m_1^2=1 \Rightarrow m= \pm 1 \quad \text { (+ ve angle) } \\ & m_1=1 \\ & m_2=\frac{3}{2} \\ \therefore & m_1+m_2=\frac{-2 h}{b} \\ \Rightarrow & 1+\frac{3}{2}=\frac{-2 h}{4} \\ \Rightarrow & h=-5 \end{array} $

Now, equation

$ 2 x^2+x y-6 y^2+k=0 $

Old equation

$ 2 x^2+x y-6 y^2-13 x+9 y+15=0 $

Origin is shifted to $(a, b)$

$ \begin{aligned} 2(x-a)^2+ & (x-a)(y-b)-6(y-b)^2 \ -13(x-a)+9(y-b)+15=0 \end{aligned} $

∴ Coefficient of $x=0$

$ 4 a+b-13=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Coefficient of $y=0$

$ a-12 b+9=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Solving Eqs. (i) and (ii) we get

$ a=3 \text { and } b=1 $

∴ Constant

$ \Rightarrow k=2 a^2+a b-6 b^2-13 a+9 b+15 $

Put $a=3$ and $b=1$

$ k=0 $

$ \text { } \begin{aligned} Point\,\,\,P & =\left(\frac{-20+12}{-5+4}, \frac{-30+20}{-5+4}\right) \\ & =\left(\frac{-8}{-1}, \frac{-10}{-1}\right)=(8,10) \end{aligned} $

$\because P$ lies on line $6 x+3 y+k=0$

$ \begin{array}{ll} \therefore \quad & 6(8)+3(10)+k=0 \\ & k=-78 \end{array} $

∴ Equation of line $6 x+3 y-78=0$

$ \Rightarrow 2 x+y-26=0 $

∴ Point of intersection of $2 x+y-26=0$ and $x-y+1=0$ is $(g, h)=(8,9)$

$ \begin{array}{ll} \therefore \quad & g-h+1=0 \\ & h=g+1 \end{array} $

$ \begin{gathered} x-1= \pm \frac{1}{2} \cos \theta \\ y-2= \pm \frac{1}{2} \sin \theta \end{gathered} $

$ \begin{aligned} &\begin{aligned} & x= \pm \frac{\cos \theta}{2}+1 \\ & y= \pm \frac{\sin \theta}{2}+2 \end{aligned}\\ &\begin{aligned} & \therefore \theta \text { lies on line } x+\sqrt{3} y-2 \sqrt{3}=0 \\ & \pm \frac{\cos \theta}{2}+1 \pm \frac{\sqrt{3}}{2} \sin \theta+2 \sqrt{3}-2 \sqrt{3}=0 \\ & \Rightarrow \quad \pm\left(\frac{\sqrt{3}}{2} \sin \theta+\frac{1}{2} \cos \theta\right)=-1 \\ & \Rightarrow \quad \sin \left(\theta+\frac{\pi}{6}\right)= \pm 1 \\ & \Rightarrow \quad \theta+\frac{\pi}{6}=\frac{\pi}{2} \text { or }-\frac{\pi}{2} \\ & \Rightarrow \quad \theta=\frac{\pi}{3} \text { or } \frac{-2 \pi}{3} \end{aligned} \end{aligned} $

$ \begin{aligned} & \left|\begin{array}{ccc} 1 & -2 & 1 \\ 2 & -3 & -1 \\ 3 & -1 & k \end{array}\right|=0 \\ & \Rightarrow 1(-3 k-1)+2(2 k+3)+1(-2+9)=0 \\ & \Rightarrow-3 k-1+4 k+6+7=0 \\ & \Rightarrow k=-12 \\ & L_1: 3 x-y+k=0 \\ & \Rightarrow 3 x-y-12=0 \Rightarrow m_1=3 \end{aligned} $

$ \begin{aligned} L_2: m x & -3 y+6=0 \\ \Rightarrow \quad m_2 & =\frac{m}{3} \\ \tan \theta & =\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ \tan 45^{\circ} & =\left|\frac{3-\frac{m}{3}}{1+m}\right| \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \quad a x^2+4 x y-2 y^2=0, h=2 a \\ & \tan \theta=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right| \\ & \Rightarrow \quad 2 \sqrt{10}=\left|\frac{2 \sqrt{4+2 a}}{a-2}\right| \\ & \Rightarrow \quad 10=\frac{4+2 a}{(a-2)^2} \\ & \Rightarrow \quad 5\left(a^2-4 a+4\right)=a+2 \\ & \Rightarrow \quad 5 a^2-21 a+18=0 \\ & \Rightarrow \quad 5 a^2-15 a-6 a+18=0 \\ & \Rightarrow \quad 5 a(a-3)-6(a-3)=0 \\ & \quad a=3, a=\frac{6}{5} \\ & \because \quad a \in Z \\ & \therefore \quad a=3 \\ & \therefore \quad m_1 m_2=\frac{a}{b}=\frac{3}{-2}=\frac{-3}{2} . \end{aligned} \end{aligned} $

Rotation by angle $\theta$ in the positive direction

$ \begin{aligned} & x^{\prime}=x \cos \theta+y \sin \theta \\ & y^{\prime}=-x \sin \theta+y \cos \theta \end{aligned} $

Let $\left(x_2, y_2\right)=(-4 \sqrt{2},-2 \sqrt{2})$

Point before rotation $\left(x_1, y_1\right)$,

$ \begin{aligned} x_1 & =x_2 \cos \left(-\frac{\pi}{4}\right)+y_2 \sin \left(-\frac{\pi}{4}\right) \\ & =x_2\left(\frac{\sqrt{2}}{2}\right)-y_2\left(\frac{\sqrt{2}}{2}\right) \end{aligned} $

And $y_1=x_2\left(\frac{\sqrt{2}}{2}\right)+y_2\left(\frac{\sqrt{2}}{2}\right)$

$ \begin{aligned} \Rightarrow \quad x_1 & =\frac{\sqrt{2}}{2}(-4 \sqrt{2})-\frac{\sqrt{2}}{2}(-2 \sqrt{2}) \\ & =-4+2=-2 \\ \text { And } y_1 & =\frac{\sqrt{2}}{2}(-4 \sqrt{2})+\frac{\sqrt{2}}{2}(-2 \sqrt{2}) \\ & =-4-2=-6 \end{aligned} $

So, point before rotation is $(-2,-6)$ reflection about $y=-x$

$\Rightarrow$ Reflection of point $(x, y)$ is $(-y,-x)$ i.e., $(6,2)$

Translation by +3 in $X$-axis

$ \begin{array}{rlrl} \therefore & (\alpha, \beta) & =(6,-3,2) \\ & & =(3,2) \\ \therefore & \alpha+\beta & =3+2=5 \end{array} $

Let slope of line passing through the point $(4,-3)$ be $m$

And slope of the line joining point $(1,1)$ and $(2,3)$ is $=\frac{3-1}{2-1}=2$

$ \begin{array}{llrl} \therefore & & \tan 45^{\circ} & =\left|\frac{m-2}{1+2 m}\right| \\ \Rightarrow & & 1 & =\left|\frac{m-2}{1+2 m}\right| \end{array} $

taking positive sign, we get

$ \begin{aligned} 1 & =\frac{m-2}{1+2 m} \\ \Rightarrow 1+2 m & =m-2 \\ \Rightarrow \quad m & =-3 \end{aligned} $

Taking negative sign, we get

$ \begin{aligned} -1 & =\frac{m-2}{1+2 m} \\ \Rightarrow \quad m-2 & =-1-2 m \Rightarrow 3 m=1 \end{aligned} $

$ \Rightarrow \quad m=\frac{1}{3} $

Since, slope is negative.

$ \therefore \quad m=-3 $

Equation of line is

$ \begin{aligned} & & y+3 & =(-3)(x-4) \\ \Rightarrow & & y+3 & =-3 x+12 \\ \Rightarrow & & 3 x+y & =9 \end{aligned} $

$\therefore x$-intercept, when $y=0$, is $3 x=9$

$ \Rightarrow \quad x=3 $

and $y$-intercept, when $x=0$ is $y=9$

$\therefore$ Sum of intercepts $=3+9=12$

We have $O(0,0), B(-3,-1), C(-1,-3)$ are vertices of a $\triangle O B C$ and equation of $D E$ is $2 x+2 y+\sqrt{2}=0$

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

$\therefore$ Slope of attitude from 0 to $B C$ is 1 $(\perp$ to $B C)$

$\therefore$ Equation of altitude with slope passes through $(0,0)$

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

Equation of $B C$

$ y+1=(-1)(x+3) \Rightarrow y=-x-4 $

Point of intersection of line $B C$ and altitude

$ \begin{aligned} & & x & =-x-4 \\ \Rightarrow & & 2 x & =-4 \\ \Rightarrow & & x & =-2 \\ \therefore & & y & =-2 \end{aligned} $

$\therefore$ Foot of perpendicular $A=(-2,-2)$

Now, $2 x+2 y+\sqrt{2}=0$

$ \begin{array}{ll} \Rightarrow & x+y=-\frac{\sqrt{2}}{2} \\ \Rightarrow & x+x=\frac{-\sqrt{2}}{2} \quad(\text { altitude } y=x) \\ \Rightarrow & 2 x=\frac{-\sqrt{2}}{2} \\ \Rightarrow & x=\frac{-\sqrt{2}}{4} \end{array} $

Also, $\quad y=\frac{-\sqrt{2}}{4}$

$\therefore$ Intersection of altitude and $D E$ is P. $\left(\frac{-\sqrt{2}}{4}, \frac{-\sqrt{2}}{4}\right)$

$ \begin{aligned} & \therefore A P=\sqrt{\left(-2+\frac{\sqrt{2}}{4}\right)^2+\left(-2+\frac{\sqrt{2}}{4}\right)^2} \\ & \quad=\sqrt{2\left(-2+\frac{\sqrt{2}}{4}\right)^2}=\sqrt{2}\left(-2+\frac{\sqrt{2}}{4}\right) \\ & \text { And } P O=\sqrt{2\left(\frac{\sqrt{2}}{4}\right)^2} \\ & \quad=\sqrt{2 \times \frac{2}{16}}=\frac{1}{2} \\ & \therefore \frac{P O}{A P}=\frac{\frac{1}{2}}{\sqrt{2}\left(-2+\frac{\sqrt{2}}{4}\right)}=1: 4 \sqrt{2}-1 \end{aligned} $

Given equation of curve $3 x+2 y-3 x y=0$ every point on the curve is the centroid of a triangle formed by the coordinate axes and a line $(L)$ Since, $L$ intersect the coordinate axes

$ \begin{aligned} \therefore \text { let } A & =(a, 0) \\ B & =(0, b) \end{aligned} $

$\therefore$ Centroid of $\triangle O A B$,

$ \begin{aligned} & (x, y)=\left(\frac{0+0+a}{3}, \frac{0+b+0}{3}\right)=\left(\frac{a}{3}, \frac{b}{3}\right) \\ & \Rightarrow a=3 x, b=3 y \end{aligned} $

Equation of line

$ \begin{aligned} & \frac{X}{3 x}+\frac{Y}{3 y}=1 \\ \Rightarrow & \frac{X}{x}+\frac{Y}{y}=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Since, centroid $(x, y)$ lies on given curve

$ \begin{aligned} & 3 x+2 y-3 x y=0 \\ & \Rightarrow \quad 3 x+2 y=3 x y \\ & \Rightarrow \quad 3 x y=3 x+2 y \Rightarrow 3=\frac{3 x}{x y}+\frac{2 y}{x y} \\ & \Rightarrow \quad \frac{X}{x}+\frac{Y}{y}=\frac{3}{y}+\frac{2}{x} \quad \text { (Using Eq. (i)) } \\ & \Rightarrow \quad \frac{X-2}{x}+\frac{Y-3}{y}=0 \end{aligned} $

$\Rightarrow$ Does not represents to a single line

Now, $3 x+2 y=3 x y$

$ \begin{aligned} & \Rightarrow \quad \frac{3 x}{3 x y}+\frac{2 y}{3 x y}=1 \\ & \Rightarrow \quad \frac{1}{y}+\frac{2}{3 x}=1 \\ & \Rightarrow \quad \frac{1}{\frac{b}{3}}+\frac{2}{3\left(\frac{a}{3}\right)}=1 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{3}{b}+\frac{2}{a}=1 \Rightarrow 3 a+2 a=a b \\ & \Rightarrow \quad a b-3 a-2 b=0 \\ & \Rightarrow \quad a=\frac{2 b}{b-3} \end{aligned} $

Consider the equation of line $L$

$ \begin{aligned} & b X+a Y=a b \\ & \Rightarrow \quad b X+\frac{2 b}{b-3} Y=b\left(\frac{2 b}{b-3}\right) \\ & \Rightarrow \quad X+\frac{2}{b-3} Y=\frac{2 b}{b-3} \\ & \Rightarrow \quad X(b-3)+2 Y=2 b \\ & \Rightarrow \quad b(X-2)+(-3 X+2 Y)=0 \end{aligned} $

For all such lines $L$

$ X-2=0 \Rightarrow X=2 $

and $\quad-3 X+2 Y=0$

$\Rightarrow \quad Y=3$

$\Rightarrow \quad L$ is passes through $(2,3)$

$\Rightarrow \quad L$ are concurrent.

Given, equation

$ \left(a^2-3\right) x^2+16 x y-2 a y^2+4 x-8 y-2=0 $

represents a pair of perpendicular line

$ \begin{aligned} & \therefore\left(a^2-3\right)+(-2 a)=0 \\ & \Rightarrow \quad a^2-2 a-3=0 \\ & \Rightarrow \quad a^2-3 a+a-3=0 \\ & \Rightarrow \quad a(a-3)+1(a-3)=0 \\ & \Rightarrow \quad(a-3)(a+1)=0 \\ & \Rightarrow \quad a=3,-1 \end{aligned} $

$\Rightarrow a=3$, since for $a=-1$ lines does not exists.

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 $3 x^2-10 x-y+10=0$