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} $

1. 1. A straight line $\boldsymbol{L}$ passes through the point of intersection of 2 straight lines: $\left\{\begin{array}{l}\boldsymbol{L}_1: 4 \boldsymbol{x}+3 \boldsymbol{y}-1=0 \\ \boldsymbol{L}_2: 3 \boldsymbol{x}+4 \boldsymbol{y}-1=0\end{array}\right.$

2. 2. The straight line $\boldsymbol{L}$ cuts the axes at $\boldsymbol{P}$ and $\boldsymbol{Q}$

3. 3. To identify the locus of the mid-point of the segment $P Q$

1. The point of intersection of $\boldsymbol{L}_1$ and $\boldsymbol{L}_2$ is $\boldsymbol{I}(\boldsymbol{\alpha}, \boldsymbol{\beta})$

2. The mid-point of the segment $\boldsymbol{P Q}$ is $\boldsymbol{M}(\boldsymbol{h}, \boldsymbol{k})$

Analytical Study :

We first solve the pair of straight lines $\boldsymbol{L}_1$ and $\boldsymbol{L}_2$ to locate the point of intersection :

$ \begin{aligned} & 4 x+3 y-1=0 \Rightarrow 12 x+9 y=3 \\ & 3 x+4 y-1=0 \Rightarrow 12 x+16 y=4 \end{aligned} \quad \Rightarrow \beta=\frac{1}{7} \Rightarrow \boldsymbol{\alpha}=\frac{1}{7} \Rightarrow \boldsymbol{I}\left(\frac{1}{7}, \frac{1}{7}\right) \ldots . . (i)$

The straight line $\boldsymbol{L}$ passes through the point $\boldsymbol{I}$ identified by equation (i)

The equation of the straight line $\boldsymbol{L}$ in point slope form is written below :

$ L:\left(y-\frac{1}{7}\right)=m\left(x-\frac{1}{7}\right) \equiv L: 7 m x-7 y=m-1 \equiv L: \frac{7 m x}{m-1}+\frac{7 y}{1-m}=1 \ldots . (ii)$

The equation (ii) gives the straight line $\boldsymbol{L}$ in intercept form where the $\boldsymbol{X}$ and $\boldsymbol{Y}$ intercepts are observed below :

From Chart 1, the mid-point of the section $\boldsymbol{P Q}$ is derived applying mid-point formulae :

Chart 2 has the $\boldsymbol{X}$ and $\boldsymbol{Y}$ coordinate each as a function of $\boldsymbol{m}$ where we can eliminate $\boldsymbol{m}$ and mathematically trace the locus :

$ \boldsymbol{k}=\frac{1-\boldsymbol{m}}{14} \Rightarrow \boldsymbol{m}=1-14 \boldsymbol{k} \Rightarrow \boldsymbol{h}=\frac{1-14 \boldsymbol{k}-1}{14(1-14 \boldsymbol{k})}=\frac{14 \boldsymbol{k}}{14(14 \boldsymbol{k}-1)}=\frac{\boldsymbol{k}}{(14 \boldsymbol{k}-1)} \Rightarrow 14 \boldsymbol{h} \boldsymbol{k}-\boldsymbol{h}=\boldsymbol{k} \Rightarrow \boldsymbol{h}+\boldsymbol{k}-14 \boldsymbol{h} \boldsymbol{k}=0 $

The locus of the mid-point of the section $\boldsymbol{P Q}$ after replacement of the dummy variables is found to be the rectangular hyperbola $\boldsymbol{x}+\boldsymbol{y}-14 \boldsymbol{x} \boldsymbol{y}=0 $

Since $QR$ lies on the line

$ x+y=4, $

and $P=(3,5)$, first find the altitude from $P$ to the side $QR$.

In an equilateral triangle, all altitudes, medians and angle bisectors coincide. Also, the orthocentre lies on the altitude from each vertex and divides the altitude in the ratio $2:1$ from the vertex.

So let us proceed step by step.

The line $QR$ has equation

$ x+y-4=0. $

Distance of point $P(3,5)$ from this line is

$ d=\frac{|3+5-4|}{\sqrt{1^2+1^2}} =\frac{4}{\sqrt{2}} =2\sqrt{2}. $

This is the altitude of the equilateral triangle from vertex $P$.

Now, the foot of the perpendicular from $P$ to the line $x+y=4$ lies on the altitude. Since in an equilateral triangle the orthocentre lies at $\frac{2}{3}$ of the altitude from the vertex, we can find it along the perpendicular direction.

The slope of $x+y=4$ is $-1$, so the perpendicular slope is $1$.

Hence the altitude through $P(3,5)$ is

$ y-5=1(x-3) $

or

$ y=x+2. $

The orthocentre lies on this line.

Now find the foot of perpendicular from $P$ to $x+y=4$ by solving

$ y=x+2 $

and

$ x+y=4. $

Substitute $y=x+2$ into $x+y=4$:

$ x+(x+2)=4 $

$ 2x=2 $

$ x=1,\quad y=3. $

So the foot is $(1,3)$.

Thus the altitude segment is from $P(3,5)$ to $(1,3)$.

The orthocentre divides this altitude in the ratio $2:1$ from vertex $P$. Therefore,

$ H=\left(\frac{2\cdot 1+1\cdot 3}{3},\frac{2\cdot 3+1\cdot 5}{3}\right) $

This is incorrect ratio placement, so let us use the correct section carefully:

If $H$ divides the segment from $P$ to foot $D$ in the ratio $2:1$, then

$ H=P+\frac{2}{3}(D-P). $

Here,

$ D-P=(1-3,\,3-5)=(-2,-2). $

So

$ H=(3,5)+\frac{2}{3}(-2,-2) =\left(3-\frac{4}{3},\,5-\frac{4}{3}\right) =\left(\frac{5}{3},\,\frac{11}{3}\right). $

Thus,

$ \alpha=\frac{5}{3},\qquad \beta=\frac{11}{3}. $

So,

$ \alpha+\beta=\frac{5}{3}+\frac{11}{3}=\frac{16}{3}. $

Therefore,

$ 9(\alpha+\beta)=9\cdot \frac{16}{3}=48. $

Hence the correct answer is

$ \boxed{48} $

So, Option D is correct.

Given:

• $L_1: x+3=0 \Rightarrow x=-3$

• $L_2: x-y=0 \Rightarrow y=x$

• $L_3: 3x+y=0 \Rightarrow y=-3x$

We first find points $A$ and $B$.

1. Coordinates of $A$

Point $A$ is the intersection of $L_1$ and $L_2$.

Since $x=-3$ and $y=x$, we get

$ y=-3 $

So,

$ A=(-3,-3) $

2. Coordinates of $B$

Point $B$ is the intersection of $L_1$ and $L_3$.

Since $x=-3$ and $y=-3x$, we get

$ y=-3(-3)=9 $

So,

$ B=(-3,9) $

3. Equation of angle bisectors of $L_2$ and $L_3$

The lines are:

$ L_2: x-y=0 $

$ L_3: 3x+y=0 $

Angle bisectors are given by

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

That is,

$ \frac{x-y}{\sqrt{2}}=\pm \frac{3x+y}{\sqrt{10}} $

We need the bisector of the obtuse angle.

Let us find its slope.

Since both lines pass through the origin, we can use the formula for slope of angle bisectors.

The slopes of $L_2$ and $L_3$ are:

• for $L_2$, $m_1=1$

• for $L_3$, $m_2=-3$

The acute angle bisector lies between the directions of these lines, and the obtuse angle bisector is the other one.

Now let us find the bisector which intersects $x=-3$ between $y=-3$ and $y=9$ or outside, by solving directly.

Substitute $x=-3$ into angle bisector equations.

4. Find point $C$

From

$ \frac{x-y}{\sqrt{2}}=\frac{3x+y}{\sqrt{10}} $

put $x=-3$:

$ \frac{-3-y}{\sqrt{2}}=\frac{-9+y}{\sqrt{10}} $

Cross-multiplying:

$ \sqrt{10}(-3-y)=\sqrt{2}(-9+y) $

This gives one point on $L_1$.

From

$ \frac{x-y}{\sqrt{2}}=-\frac{3x+y}{\sqrt{10}} $

put $x=-3$:

$ \frac{-3-y}{\sqrt{2}}=-\frac{-9+y}{\sqrt{10}} $

Instead of solving both, it is easier to use the fact that any point on the angle bisector is equidistant from the two lines.

Let $C=(-3,k)$ on $L_1$.

Distance of $C$ from $L_2: x-y=0$ is

$ \frac{|x-y|}{\sqrt{2}}=\frac{|-3-k|}{\sqrt{2}}=\frac{|k+3|}{\sqrt{2}} $

Distance of $C$ from $L_3: 3x+y=0$ is

$ \frac{|3x+y|}{\sqrt{10}}=\frac{|-9+k|}{\sqrt{10}}=\frac{|k-9|}{\sqrt{10}} $

So,

$ \frac{|k+3|}{\sqrt{2}}=\frac{|k-9|}{\sqrt{10}} $

Squaring:

$ \frac{(k+3)^2}{2}=\frac{(k-9)^2}{10} $

$ 5(k+3)^2=(k-9)^2 $

Expand:

$ 5(k^2+6k+9)=k^2-18k+81 $

$ 5k^2+30k+45=k^2-18k+81 $

$ 4k^2+48k-36=0 $

$ k^2+12k-9=0 $

$ k=-6\pm 3\sqrt{5} $

So the two bisectors meet $L_1$ at points with $y=-6+3\sqrt{5}$ and $y=-6-3\sqrt{5}$.

Now $A=(-3,-3)$ and $B=(-3,9)$.

The obtuse angle is the larger angle between $L_2$ and $L_3$.

Its bisector lies inside that larger region, which is the bisector with slope closer to negative small value, and it cuts $x=-3$ at the point between $A$ and $B$.

Since

$ -6+3\sqrt{5}\approx 0.708 $

this lies between $-3$ and $9$.

Hence,

$ C=(-3,\,-6+3\sqrt{5}) $

5. Find $AC^2$ and $BC^2$

Since all three points have the same $x$-coordinate, distances depend only on $y$-coordinates.

$AC$

$ AC=|-6+3\sqrt{5}-(-3)|=|-3+3\sqrt{5}|=3(\sqrt{5}-1) $

So,

$ AC^2=9(\sqrt{5}-1)^2 $

$BC$

$ BC=|9-(-6+3\sqrt{5})|=|15-3\sqrt{5}|=3(5-\sqrt{5}) $

So,

$ BC^2=9(5-\sqrt{5})^2 $

Therefore,

$ BC^2:AC^2=(5-\sqrt{5})^2:(\sqrt{5}-1)^2 $

Now,

$ (5-\sqrt{5})^2=25+5-10\sqrt{5}=30-10\sqrt{5}=10(3-\sqrt{5}) $

and

$ (\sqrt{5}-1)^2=5+1-2\sqrt{5}=6-2\sqrt{5}=2(3-\sqrt{5}) $

Hence,

$ BC^2:AC^2=10(3-\sqrt{5}):2(3-\sqrt{5})=5:1 $

So the required ratio is

$ \boxed{5:1} $

Hence, the correct option is $\boxed{\text{A}}$.

Let the coordinates of vertex B be $(x_B, y_B)$. We are given that vertex A is $(1, 2)$ and the mid-point of side AB is $(5, -1)$.

Using the mid-point formula, the coordinates of the mid-point are given by:

$ \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right) = (5, -1) $

By equating the $x$ and $y$ coordinates, we get:

$ \frac{1 + x_B}{2} = 5 \implies 1 + x_B = 10 \implies x_B = 9 $

$ \frac{2 + y_B}{2} = -1 \implies 2 + y_B = -2 \implies y_B = -4 $

So, the coordinates of vertex B are $(9, -4)$.

Let the coordinates of vertex C be $(x_C, y_C)$. The centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

$ \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $

We are given the centroid of triangle ABC is $(3, 4)$. Using the coordinates of vertices A, B, and C:

$ \frac{1 + 9 + x_C}{3} = 3 \implies 10 + x_C = 9 \implies x_C = -1 $

$ \frac{2 + (-4) + y_C}{3} = 4 \implies -2 + y_C = 12 \implies y_C = 14 $

So, the coordinates of vertex C are $(-1, 14)$.

The circumcenter of a triangle is a point that is equidistant from all three of its vertices. Let the given circumcenter be marked as $O(\alpha, \beta)$. Therefore, we have:

$ OA = OB = OC \implies OA^2 = OB^2 = OC^2 $

Using the distance formula, we will write the equations for the squares of the distances:

$ OA^2 = (\alpha - 1)^2 + (\beta - 2)^2 = \alpha^2 + \beta^2 - 2\alpha - 4\beta + 5 $

$ OB^2 = (\alpha - 9)^2 + (\beta + 4)^2 = \alpha^2 + \beta^2 - 18\alpha + 8\beta + 97 $

$ OC^2 = (\alpha + 1)^2 + (\beta - 14)^2 = \alpha^2 + \beta^2 + 2\alpha - 28\beta + 197 $

Now, equating $OA^2 = OB^2$:

$ \alpha^2 + \beta^2 - 2\alpha - 4\beta + 5 = \alpha^2 + \beta^2 - 18\alpha + 8\beta + 97 $

$ 16\alpha - 12\beta = 92 $

Dividing the entire equation by 4 gives our first linear equation:

$ 4\alpha - 3\beta = 23 \quad \text{--- (Equation 1)} $

Next, equating $OA^2 = OC^2$:

$ \alpha^2 + \beta^2 - 2\alpha - 4\beta + 5 = \alpha^2 + \beta^2 + 2\alpha - 28\beta + 197 $

$ -4\alpha + 24\beta = 192 $

Dividing this equation by -4 gives our second linear equation:

$ \alpha - 6\beta = -48 \implies \alpha = 6\beta - 48 \quad \text{--- (Equation 2)} $

Substitute the expression for $\alpha$ from Equation 2 into Equation 1:

$ 4(6\beta - 48) - 3\beta = 23 $

$ 24\beta - 192 - 3\beta = 23 $

$ 21\beta - 192 = 23 $

$ 21\beta = 215 \implies \beta = \frac{215}{21} $

Now, substitute the value of $\beta$ back into Equation 2 to find $\alpha$:

$ \alpha = 6\left(\frac{215}{21}\right) - 48 $

$ \alpha = \frac{1290}{21} - \frac{1008}{21} $

$ \alpha = \frac{282}{21} $

The problem asks for the value of $21(\alpha + \beta)$. We can simply plug in our values:

$ 21(\alpha + \beta) = 21 \left( \frac{282}{21} + \frac{215}{21} \right) $

$ 21(\alpha + \beta) = 282 + 215 = 497 $

Option C is the correct answer.

Let the midpoints of the sides of triangle $ABC$ be

$ D\left(\frac{5}{2}, 7\right), \quad E\left(\frac{5}{2}, 3\right), \quad F(4, 5) $

We have to find the incentre $(h,k)$ of triangle $ABC$.

Step 1: Use the fact about the medial triangle

The triangle formed by joining the midpoints of the sides of a triangle is called the medial triangle.

If the midpoints are known, then the original triangle can be reconstructed using:

$ A = E + F - D,\quad B = F + D - E,\quad C = D + E - F $

Now calculate each vertex.

Vertex $A$

$ A = \left(\frac{5}{2}, 3\right) + (4, 5) - \left(\frac{5}{2}, 7\right) $

$ A = (4, 1) $

Vertex $B$

$ B = (4, 5) + \left(\frac{5}{2}, 7\right) - \left(\frac{5}{2}, 3\right) $

$ B = (4, 9) $

Vertex $C$

$ C = \left(\frac{5}{2}, 7\right) + \left(\frac{5}{2}, 3\right) - (4, 5) $

$ C = (1, 5) $

So the triangle has vertices

$ A(4,1),\quad B(4,9),\quad C(1,5) $

Step 2: Find the side lengths

Using distance formula:

Side opposite $A$, i.e. $BC$

$ BC = \sqrt{(4-1)^2 + (9-5)^2} = \sqrt{3^2 + 4^2} = 5 $

Side opposite $B$, i.e. $CA$

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

Side opposite $C$, i.e. $AB$

$ AB = \sqrt{(4-4)^2 + (9-1)^2} = 8 $

Thus,

$ a = 5,\quad b = 5,\quad c = 8 $

Step 3: Use incentre formula

The incentre of triangle with vertices $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$ is

$ \left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c} \right) $

where

• $a = BC$

• $b = CA$

• $c = AB$

Substitute the values:

$ h = \frac{5\cdot 4 + 5\cdot 4 + 8\cdot 1}{5+5+8} = \frac{20+20+8}{18} = \frac{48}{18} = \frac{8}{3} $

$ k = \frac{5\cdot 1 + 5\cdot 9 + 8\cdot 5}{18} = \frac{5+45+40}{18} = \frac{90}{18} = 5 $

So the incentre is

$ \left(\frac{8}{3}, 5\right) $

Step 4: Find $3h + k$

$ 3h + k = 3\cdot \frac{8}{3} + 5 = 8 + 5 = 13 $

Therefore, the answer is

$ \boxed{13} $

So, the correct option is Option C.

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

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

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

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

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

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