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} $

$\begin{aligned} & \frac{x}{a}+\frac{y}{b}=1 \\ & \frac{3}{a}+\frac{5}{b}=1 \\ & 3 b+5 a=a b \\ & 5 a-a b=-3 b \\ & a(5-b)=-3 b \\ & a=\frac{3 b}{b-5} \end{aligned}$

$\begin{aligned} & \text { Area of triangle }=\left|\frac{1}{2} \times a \times b\right| \\ & =\frac{1}{2} \times \frac{3 b}{b-5} \times b \\ & \Rightarrow f(b)=\frac{3 b^2}{2 b-10} \\ & \Rightarrow f^{\prime}(b)=0,(2 b-10) 6 b-2\left(3 b^2\right)=0 \\ & 12 b^2-60 b-6 b^2=0 \\ & 6 b^2-60 b=0 \\ & b^2-10 b=0 \\ & b(b-10)=0 \\ & b=0 \text { or } b=10 \end{aligned}$

So for minimum area, $b=10$

then $\frac{1}{2} \times a \times b=30$

$\begin{aligned} & P^{\prime} R: y+2=\frac{5}{3}(x-1) \\ & \text { For Point } Q \Rightarrow y=0 \\ & \frac{6}{5}=a-1 \Rightarrow a=\frac{11}{5} \end{aligned}$

Now, $P Q R S$ is parallelogram

$\begin{aligned} & \therefore \frac{h+a}{2}=\frac{4+1}{2} \Rightarrow h=5-\frac{11}{5}=\frac{14}{5} \\ & \text { and } \frac{2+3}{2}=\frac{k}{2} \Rightarrow K=5 \end{aligned}$

Now $h k^2=25 \times \frac{14}{5}=14 \times 5=70$

To solve this problem, we will use the concept of the slope and tangent of the angle subtended by the line segments at the origin.

Given points: $(5,2)$ and $(2,a)$

The slope of the line joining the origin and $(5,2)$ is:

The slope of the line joining the origin and $(2,a)$ is:

The line segment joining these points subtends an angle $\theta = \frac{\pi}{4}$ at the origin.

According to the tangent formula for the angle between two lines:

Here, $\theta = \frac{\pi}{4}$ so, $\tan \left( \frac{\pi}{4} \right) = 1$. Therefore,

This simplifies to:

Multiplying both the numerator and denominator by 10 to simplify:

This leads to two equations due to the absolute value:

1. $ 5a - 4 = 10 + 2a $

2. $ 5a - 4 = -(10 + 2a) $

Thus, the possible values of $a$ are $\frac{14}{3}$ and $-\frac{6}{7}$.

The product of these values is:

Therefore, the absolute value of the product of all possible values of $a$ is 4.

Answer: Option A

$\begin{aligned} & 2=\frac{2 x_2+x_1}{3}, \frac{-4}{3}=\frac{2 y_2+y_1}{3} \\ & 2 x_2+x_1=6,2 y_2+y_1=-4 \end{aligned}$

$\begin{aligned} & x_1=6-2 x_2 \quad \text{.... (1)}\\ & y_1=-4-2 y_2 \quad \text{.... (2)}\\ & 4 x_1+y_1=14 \quad \text{.... (3)}\\ & 3 x_2-2 y_2=5 \quad \text{.... (4)} \end{aligned}$

From here, $x_2=1, y_2=-1, x_1=4, y_1=-2$

$\begin{aligned} & B(4,-2) C(1,-1) \\ & y+2=\frac{-1+2}{1-4}(x-4) \\ & -3 y-6=x-4 \\ & x+3 y+2=0 \end{aligned}$

$\begin{aligned} & \therefore \frac{(h-2)^2+(k-1)^2}{(h-1)^2+(k-3)^2}=\frac{25}{16} \\ & \frac{h^2+k^2-4 h-2 k+5}{h^2+k^2-2 h-6 k+10}=\frac{25}{16} \\ \end{aligned}$

Replacing $h \rightarrow x$ and $k \rightarrow y$.

$\begin{aligned} & 16 x^2+16 y^2-64 x-32 y+80 \\ & =25 x^2+25 y^2-50 x-150 y+250 \\ & 9 x^2+9 y^2+14 x-118 y+170=0 \\ & \Rightarrow a=9, b=9, c=0, d=14, e=-118 \\ & a^2+2 b+3 c+4 d+e \\ & 81+18+56-118 \\ & \Rightarrow 37 \end{aligned}$

$\begin{aligned} & L:(y+9)=m(x-4) \\ & A:\left(4+\frac{9}{m}, 0\right) \\ & B:(0,-9-4 m) \\ & O A+O B]_{\min } \\ & \Rightarrow E_{\min }=4+\frac{9}{m}+9+4 m \\ & E=13+\frac{9}{m}+4 m \\ & \frac{d E}{d M}=0 \Rightarrow-\frac{9}{m^2}+4=0 \Rightarrow m= \pm \frac{3}{2} \end{aligned}$

$\begin{aligned} & \frac{d^2 E}{d M^2}=\frac{18}{m^3}>0 \text { for } \quad m=\frac{3}{2} \\ & \therefore E_{\min }=4+6+9+6=25 \end{aligned}$

$O P=O Q$ ($\triangle P Q R$ is isosceles triangle)

Let slope of line $O P \rightarrow m_1$

So, equation $\rightarrow y=m_1 x$

$\begin{aligned} &\begin{aligned} & \tan 45^{\circ}=\left|\frac{m_1-m_2}{m_1 m_2}\right| \\ & 1=\left|\frac{m_1+\frac{3}{4}}{1-\frac{3}{4} m_1}\right| \\ & \Rightarrow 1-\frac{3}{4} m_1=m_1+\frac{3}{4} \\ & \frac{1}{4}=\frac{7}{4} m_1 \Rightarrow m_1=\frac{1}{7} \end{aligned}\\ &\text { Equation } O P \rightarrow y=\frac{1}{7} x \end{aligned}$

Point of intersection of $O P$ & line $3 x+4 y=12$ is $P\left(\frac{84}{25}, \frac{12}{25}\right)$

$\begin{aligned} & \Rightarrow \quad O P^2=a^2=\left(\frac{84}{25}\right)^2+\left(\frac{12}{25}\right)^2=\frac{288}{25} \\ & \therefore \quad I=O P^2+P Q^2+Q O^2 \\ &=a^2+a^2+2 a^2 \\ &=4 a^2 \\ &=4 \times \frac{288}{25} \\ & I=46.08 \\ & {[I] }=46 \end{aligned}$

Equation of $A C: x+y=2$

Equation of $A B: x-y+4=0$

Equation of $B C: 3 x+5 y=4$

The line nearest to origin is parallel to $A C$ and inward. Let its equation is $x+y=C$.

$\therefore\left|\frac{C-2}{\sqrt{2}}\right|=1$

$\therefore \quad C=2-\sqrt{2}$

$\therefore$ required equation line is :

$x+y-(2-\sqrt{2})=0$

$\mathrm{A}(\mathrm{a}, \mathrm{b}), \quad \mathrm{B}(3,4), \quad \mathrm{C}(-6,-8)$

$\Rightarrow \mathrm{a}=0, \mathrm{~b}=0 \quad \Rightarrow \mathrm{P}(3,5)$

Distance from $\mathrm{P}$ measured along $\mathrm{x}-2 \mathrm{y}-1=0$

$\Rightarrow x=3+r \cos \theta, \quad y=5+r \sin \theta$

$\begin{aligned} & \text { Where } \tan \theta=\frac{1}{2} \\ & \mathrm{r}(2 \cos \theta+3 \sin \theta)=-17 \\ & \Rightarrow \mathrm{r}=\left|\frac{-17 \sqrt{5}}{7}\right|=\frac{17 \sqrt{5}}{7} \end{aligned}$

Let E is mid point of diagonals

$\frac{\alpha+\gamma}{2}=\frac{1+1}{2}$

$\alpha+\gamma=2$

& $\frac{\beta+\delta}{2}=\frac{2+0}{2}$

$\beta+\delta=2$

$2(\alpha+\beta+\gamma+\delta)=2(2+2)=8$

Cocus of point $\mathrm{P}(\mathrm{x}, \mathrm{y})$ whose distance from Gives $X+2 y+7=0$ & $2 x-y+8=0$ are equal is $\frac{x+2 y+7}{\sqrt{5}}= \pm \frac{2 x-y+8}{\sqrt{5}}$

$(x+2 y+7)^2-(2 x-y+8)^2=0$

Combined equation of lines

$\begin{aligned} & (x-3 y+1)(3 x+y+15)=0 \\ & 3 x^2-3 y^2-8 x y+18 x-44 y+15=0 \\ & x^2-y^2-\frac{8}{3} x y+6 x-\frac{44}{3} y+5=0 \\ & x^2-y^2+2 h x y+2 g x 2+2 f y+c=0 \\ & h=\frac{4}{3}, g=3, f=-\frac{22}{3}, c=5 \\ & g+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}=8+6=14 \end{aligned}$

$\mathrm{Eq}^{\mathrm{n}}: y-0=\tan 15^{\circ}(x-9) \Rightarrow y=(2-\sqrt{3})(x-9)$

Solving lines $\mathrm{L}_1(3 \mathrm{x}+2 \mathrm{y}=14)$ and $\mathrm{L}_2(5 \mathrm{x}-\mathrm{y}=6)$ to get $\mathrm{A}(2,4)$ and solving lines $\mathrm{L}_3(4 \mathrm{x}+3 \mathrm{y}=8)$ and $\mathrm{L}_4(6 \mathrm{x}+\mathrm{y}=5)$ to get $\mathrm{B}\left(\frac{1}{2}, 2\right)$.

Finding Eqn. of $\mathrm{AB}: 4 \mathrm{x}-3 \mathrm{y}+4=0$

Calculate distance PM

$\Rightarrow\left|\frac{4(5)-3(-2)+4}{5}\right|=6$

Writing $P$ in terms of parametric co-ordinates $2+r$

$\begin{aligned} & \cos \theta, 3+\mathrm{r} \sin \theta \text { as } \tan \theta=\sqrt{3} \\ & \mathrm{P}\left(2+\frac{\mathrm{r}}{2}, 3+\frac{\sqrt{3} \mathrm{r}}{2}\right) \end{aligned}$

$\mathrm{P}$ must satisfy $2 \mathrm{x}-3 \mathrm{y}+28=0$

So, $2\left(2+\frac{r}{2}\right)-3\left(3+\frac{\sqrt{3} \mathrm{r}}{2}\right)+28=0$

We find $r=4+6 \sqrt{3}$

$\begin{aligned} & A D: D C=1: 2 \\ & \frac{4-\alpha}{6-\alpha}=\frac{10}{8} \\ & \alpha=\beta \\ & \alpha=14 \text { and } \beta=14 \end{aligned}$

$\begin{aligned} & \mathrm{P}\left(\mathrm{a}^2, \mathrm{a}+1\right) \\ & \mathrm{L}_1=3 \mathrm{x}-\mathrm{y}+1=0 \end{aligned}$

Origin and $\mathrm{P}$ lies same side w.r.t. $\mathrm{L}_1$

$\begin{aligned} & \Rightarrow \mathrm{L}_1(0) \cdot \mathrm{L}_1(\mathrm{P})>0 \\ & \therefore 3\left(\mathrm{a}^2\right)-(\mathrm{a}+1)+1>0 \end{aligned}$

$\begin{aligned} & \Rightarrow 3 \mathrm{a}^2-\mathrm{a}>0 \\ & \mathrm{a} \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right) \quad \text{..... (1)} \end{aligned}$

Let $L_2: x+2 y-5=0$

Origin and $\mathrm{P}$ lies same side w.r.t. $\mathrm{L}_2$

$\begin{aligned} & \Rightarrow \mathrm{L}_2(0) \cdot \mathrm{L}_2(\mathrm{P})>0 \\ & \Rightarrow \mathrm{a}^2+2(\mathrm{a}+1)-5<0 \\ & \Rightarrow \mathrm{a}^2+2 \mathrm{a}-3<0 \\ & \Rightarrow(\mathrm{a}+3)(\mathrm{a}-1)<0 \end{aligned}$

$\therefore \mathrm{a} \in(-3,1)\quad \text{..... (2)}$

Intersection of (1) and (2)

$\mathrm{a} \in(-3,0) \cup\left(\frac{1}{3}, 1\right)$

Co-ordinates of $\mathrm{A}=\left(\frac{5}{3}, \frac{8}{3}\right)$

Co-ordinates of $\mathrm{B}=\left(\frac{10}{3}, \frac{4}{3}\right)$

Slope of $\mathrm{OA}=\mathrm{m}_1=\frac{8}{5}$

Slope of $\mathrm{OB}=\mathrm{m}_2=\frac{2}{5}$

$\begin{aligned} & \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ & \tan \theta=\frac{\frac{6}{5}}{1+\frac{16}{25}}=\frac{30}{41} \\ & \tan \theta=\frac{30}{41} \end{aligned}$

1. Solve the equations $15x - y = 82$ and $6x - 5y = -4$

Multiply the first equation by 5 and the second by 1 and then subtract the second from the first:

$75x - 5y = 410$

$6x - 5y = -4$

Subtracting these gives $69x = 414$ which leads to $x = 6$

Substitute $x = 6$ into the first equation to get $y = 8$

So, the first vertex is $(6,8)$.

1. Solve the equations $15x - y = 82$ and $9x + 4y = 17$

Multiply the first equation by 4 and the second by 1 and then add:

$60x - 4y = 328$

$9x + 4y = 17$

Adding these gives $69x = 345$ which leads to $x = 5$

Substitute $x = 5$ into the second equation to get $y = -7$

So, the second vertex is $(5,-7)$.

1. Solve the equations $6x - 5y = -4$ and $9x + 4y = 17$

Multiply the first equation by 4 and the second by 5 and then add:

$24x - 20y = -16$

$45x + 20y = 85$

Adding these gives $69x = 69$ which leads to $x = 1$

Substitute $x = 1$ into the first equation to get $y = 2$

So, the third vertex is $(1,2)$.

So, the vertices of the triangle are $(6,8)$, $(5,-7)$, and $(1,2)$.

Now that we have the vertices of the triangle, we can find the centroid. The centroid $(\alpha, \beta)$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by :

$ \alpha=\frac{x_1+x_2+x_3}{3}$,

$ \beta=\frac{y_1+y_2+y_3}{3} $

Substituting the coordinates of the vertices into these equations, we get:

$ \alpha=\frac{6+5+1}{3}=4 $,

$ \beta=\frac{8-7+2}{3}=1 $

Then, we find $\alpha+2\beta$ and $2\alpha-\beta$:

$ \alpha+2\beta=4+2(1)=6 \ 2\alpha-\beta=2(4)-1=7 $

So, $\alpha+2\beta=6$ and $2\alpha-\beta=7$ are the roots of the quadratic equation.

We can write the quadratic equation as

$x^2-(\alpha+2\beta+2\alpha-\beta)x+(\alpha+2\beta)(2\alpha-\beta)=0$

Substituting $\alpha=4$, $\beta=1$ gives us:

$x^2-13x+42=0$