Straight Lines and Pair of Straight Lines

Let $B$ and $D$ be other two coordinates of the square lying on the line

$ \begin{array}{r} 7 x-y+8=0 \\ B\left(x_1, 7 x_1+8\right) \\ D\left(x_2, 7 x_2+8\right) \end{array} $

Mid-point of $B$ and $D$ is $\left(-\frac{1}{2}, \frac{9}{2}\right)$.

$ \begin{array}{ll} \therefore & \frac{x_1+x_2}{2}=\frac{-1}{2} \\ \Rightarrow & x_1+x_2=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Also,

$ \begin{aligned} & (A M)^2=(B M)^2 \\ & \Rightarrow\left(-4+\frac{1}{2}\right)^2+\left(5-\frac{9}{2}\right)^2=\left(x_1+\frac{1}{2}\right)^2 +\left(7 x_1+8-\frac{9}{2}\right)^2 \\ & \Rightarrow \frac{49}{4}+\frac{1}{4}=\left(x_1+\frac{1}{2}\right)^2+49\left(x_1+\frac{1}{2}\right)^2 \\ & \Rightarrow 50\left(x_1+\frac{1}{2}\right)^2=\frac{50}{4} \\ & \Rightarrow \quad x_1+\frac{1}{2}= \pm \frac{1}{2} \\ & x_1=0,-1 \\ & x_1=0, y_1=8 \\ & x_1=-1, y_1=1 \end{aligned} $

∴ Required coordinates of the two vertices are $(0,8),(-1,1)$.

Let $P(\alpha, 2 \alpha-1)$ and its image with respect to the line $3 x-2 y+4=0$ is $Q(h, k)$

$ \begin{aligned} & \begin{aligned} \Rightarrow \frac{h-\alpha}{3} & =\frac{k-(2 \alpha-1)}{-2} \\ & =-2\left(\frac{3 \alpha-2(2 \alpha-1)+4}{3^2+(-2)^2}\right) \\ \Rightarrow \frac{h-\alpha}{3} & =\frac{k-(2 \alpha-1)}{-2}=\frac{2 \alpha-12}{13} \\ \Rightarrow \quad h & =\alpha+\frac{6 \alpha-36}{13}=\frac{19 \alpha-36}{13} \\ \text { and } \quad k & =2 \alpha-1+\frac{24-4 \alpha}{13}=\frac{22 \alpha+11}{13} \\ \Rightarrow \quad \alpha & =\frac{13 h+36}{19}, \alpha=\frac{13 k-11}{22} \end{aligned} \end{aligned} $

So, $\quad \frac{13 h+36}{19}=\frac{13 k-11}{22}$

$ \begin{aligned} & \Rightarrow \quad 22(13 h+36)=19(13 k-11) \\ & \text { Locus: } 22(13 x+36)=19(13 y-11) \end{aligned} $

Let the foot of perpendicular be $M(x, y)$.

$ \begin{aligned} & \frac{x-5}{3}=\frac{y+7}{-5}=\frac{-(3 \times 5-5 \times(-7)+1)}{3^2+(-5)^2} \\ & \frac{x-5}{3}=\frac{y+7}{-5}=\frac{51}{34}=-\frac{3}{2} \\ & \Rightarrow x=5-\frac{9}{2}=\frac{1}{2}, y=\frac{15}{2}-7=\frac{1}{2} \end{aligned} $

So, $M\left(\frac{1}{2}, \frac{1}{2}\right)$

Now perpendicular distance of $M$ to the line $2 x+5 y-3=0$ is given by

$ \begin{aligned} &\text { ∴ Perpendicular distance }\\ &=\left|\frac{2 \times \frac{1}{2}+5 \times \frac{1}{2}-3}{\sqrt{4+25}}\right|=\frac{1}{2 \sqrt{29}} \end{aligned} $

Given, $A(-1,3), B(3,5), C(5,7)$.

Let $P(x, y)$ is equidistant from $A, B, C$.

So,  $ P A^2=P B^2 $

$ \begin{aligned} & (x+1)^2+(y-3)^2=(x-3)^2+(y-5)^2 \\ & \Rightarrow x^2+2 x+1+y^2-6 y+9 \\ & \quad=x^2-6 x+9+y^2-10 y+25 \end{aligned} $

$ \begin{array}{lrl} \Rightarrow & & 8 x+4 y=24 \\ \Rightarrow & & 2 x+y=6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { Again, } & P A^2=P C^2 \end{array} $

$ \begin{aligned} &\begin{array}{lrl} \Rightarrow & (x+1)^2+(y-3)^2 & =(x-5)^2+(y-7)^2 \\ \Rightarrow & 12 x+8 y & =64 \\ \Rightarrow & 3 x+2 y & =16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{array}\\ &\text { From Eqs. (i) and (ii), } x=-4, y=14\\ &\begin{aligned} \therefore P(-4,14) & \\ \text { Hence, } P A & =\sqrt{(-4+1)^2+(14-3)^2} \\ & =\sqrt{9+121}=\sqrt{130} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { (A) } 5 x^2+2 \sqrt{7} x y-y^2=0\\ &\begin{aligned} & \tan \theta=\frac{2 \sqrt{h^2-a b}}{a+b} \\ & =\frac{2 \sqrt{(\sqrt{7})^2-5 \times(-1)}}{5-1}=\frac{2 \sqrt{12}}{4} \\ & =\sqrt{3} \\ & \Rightarrow \cos \theta=\frac{1}{2} \\ & \text { } \begin{aligned}(B)\, & x^2+\sqrt{11} x y+2 y^2=0 \\ & \tan \theta=\frac{2 \sqrt{\left(\sqrt{11} / 2^2-1 \times 2\right.}}{1+2} \\ &= \frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}} \Rightarrow \cos \theta=\frac{\sqrt{3}}{2} \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { (C) } x^2+2 \sqrt{2} x y+y^2=0 \\ & \Rightarrow \tan \theta=\frac{2 \sqrt{(\sqrt{2})^2-1 \times 1}}{1+1}=1 \\ & \Rightarrow \cos \theta=\frac{1}{\sqrt{2}} \\ & \text { (D) } 3 x^2+4 \sqrt{2} x y+y^2=0 \\ & \tan \theta=\frac{2 \sqrt{(2 \sqrt{2})^2-3 \times 1}}{3+1}=\frac{\sqrt{5}}{2} \\ & \Rightarrow \cos \theta=\frac{2}{3} \end{aligned} $

Given, $a x^2+6 x y-2 y^2=0$ represent a pair of perpendicular lines.

$ \begin{array}{ll} \text { So, } & m_1 m_2=-1 \\ \Rightarrow & \frac{a}{-2}=-1 \\ \Rightarrow & a=2 \end{array} $

Given, $9 x^2+2 h x y+4 y^2=0$ represent a pair of coincident lines.

$ \begin{array}{ll} \therefore & 4\left(\frac{y}{x}\right)^2+2 h\left(\frac{y}{x}\right)+9=0 \\ \Rightarrow & 4 m^2+2 h m+9=0 \end{array} $

For coincident lines, $D=0$

$ \begin{aligned} \Rightarrow & & 4 h^2-4 \times 4 \times 9 & =0 \\ \Rightarrow & & h & =3 \times 2 \\ \therefore& & h & =3 a \end{aligned} $

Line $x+2 y=k$ meets the curve $2 x^2-2 x y+3 y^2+2 x-y-1=0$ at $A$ and $B$.

So, combined equation of $O A$ and $O B$ is

$ \begin{gathered} 2 x^2-2 x y+3 y^2+2 x\left(\frac{x+2 y}{k}\right) \\ -y\left(\frac{x+2 y}{k}\right)-\left(\frac{x+2 y}{k}\right)^2=0 \\ \Rightarrow 2 x^2-2 x y+3 y^2 \\ +\frac{2 x^2+4 x y-y x-2 y^2}{k} \\ -\left(\frac{x^2+4 y^2+4 x y}{k^2}\right)=0 \end{gathered} $

$ \begin{aligned} &\begin{aligned} \left(2 k^2+2 k-1\right) x+ & \left(-2 k^2+3 k-4\right) x y \\ & +\left(3 k^2-2 k-4\right) y^2=0 \end{aligned}\\ &\text { For } O A \text { and } O B \text { perpendicular }\\ &\begin{array}{lcl} & & a+b=0 \\ \Rightarrow & & 2 k^2+2 k-1+3 k^2-2 k-4=0 \\ \Rightarrow & & 5 k^2-5=0 \\ \Rightarrow & & k= \pm 1 \end{array} \end{aligned} $

Let the line $L$ intersect $X$-axis at a point $(h, 0)$.

Slope of line $L\left(m_1\right)=\frac{0+3}{h-5}=\frac{3}{h-5}$

Slope of line $\left(m_2\right), \sqrt{3} x+y-9=0$ is $-\sqrt{3}$

$ \begin{aligned} & \tan 60^{\circ}=\left|\frac{\frac{3}{h-5}+\sqrt{3}}{1-\frac{3 \sqrt{3}}{h-5}}\right| \\ & \Rightarrow \quad \sqrt{3}=\left|\frac{3+\sqrt{3}(h-5)}{h-5-3 \sqrt{3}}\right| \\ & \Rightarrow \quad \sqrt{3}= \pm \frac{3+\sqrt{3}(h-5)}{h-5-3 \sqrt{3}} \\ & \text { When } \sqrt{3}=\frac{3+\sqrt{3}(h-5)}{h-5-3 \sqrt{3}} \\ & \Rightarrow \quad \sqrt{3} h-5 \sqrt{3}-9=3+\sqrt{3} h-5 \sqrt{3} \end{aligned} $

(Not possible)

When $\sqrt{3}=\frac{-(3+\sqrt{3}(h-9)}{h-5-3 \sqrt{3}}$

$ \begin{array}{llrl} \Rightarrow & & \sqrt{3} h-5 \sqrt{3}-9 & =-3-\sqrt{3} h+5 \sqrt{3} \\ \Rightarrow & & 2 \sqrt{3} h & =10 \sqrt{3}+6 \\ \Rightarrow & & h & =5+\sqrt{3} \end{array} $

The line $L$ intersect $X$-axis at point $(5+\sqrt{3}, 0)$.

Equation of line $L$ is

$ \begin{array}{ll} & y+3=\frac{3}{\sqrt{3}}(x-5) \\ \Rightarrow & y+3=\sqrt{3}(x-9) \\ \Rightarrow & \sqrt{3} x-y-5 \sqrt{3}-3=0 \end{array} $

Given, $a \alpha+\beta b+\gamma c=0$

$ \Rightarrow \quad b=-\frac{a \alpha-\gamma c}{\beta} $

The equation which represent the family of concurrent straight line is

$ \begin{aligned} & a x+b y+c=0 \\ & \quad a x+\left(\frac{-a \alpha-\gamma c}{\beta}\right) y+c=0 \\ & \Rightarrow a\left(x-\frac{\alpha y}{\beta}\right)+c\left(1-\frac{\gamma y}{\beta}\right)=0 \end{aligned} $

It represent a family of lines passing through the intersect of both the line

$ \begin{aligned} &\begin{array}{rlrl} \Rightarrow & \gamma y / \beta =1 \\ \Rightarrow & y =\beta / \gamma \\ \text { Also, } & x-\frac{\alpha y}{\beta} =0 \\ \Rightarrow & x-\frac{\alpha}{\beta}\left(\frac{\beta}{\gamma}\right) =0 \\ & x =\alpha / \gamma \end{array}\\ &\text { ∴ The required point is }\left(\frac{\alpha}{\gamma}, \frac{\beta}{\gamma}\right) \text {. } \end{aligned} $

Let the equation of line be

$ a x+b y+c=0 $

Sum of distance from $(2,0),(0,2)$ and $(1,1)$ is 0

$ \begin{aligned} & \frac{2 a+c}{\sqrt{a^2+b^2}}+\frac{2 b+c}{\sqrt{a^2+b^2}}+\frac{a+b+c}{\sqrt{a^2+b^2}}=0 \\ \Rightarrow \quad & 3 a+3 b+3 c=0 \\ \Rightarrow \quad & a+b+c=0 \end{aligned} $

For the line $a x+b y+c=0$, if $a+b+c=0$, then the line passes through the point $(1,1)$.

Given,

$ \begin{gathered} 6 a^2-3 b^2-c^2+7 a b-a c+4 b c=0 \\ \Rightarrow c^2+c(a-4 b)+3 b^2-6 a^2-7 a b=0 \end{gathered} $

which is the quadratic equation, from the quadratic formula,

$ \begin{aligned} c & =\frac{4 b-a \pm \sqrt{\begin{array}{c} a^2+16 b^2-8 a b-12 b^2 \\ +24 a^2+28 a b \end{array}}}{2} \\ & =\frac{4 b-a \pm \sqrt{25 a^2+4 b^2+20 a b}}{2} \\ c & =\frac{(4 b-a) \pm(5 a+2 b)}{2} \end{aligned} $

$ \begin{array}{lrl} \Rightarrow & & 2 c=4 b-a+5 a+2 b \\ \text { and } & & 2 c=4 b-a-5 a-2 b \\ \Rightarrow & & 4 a+6 b-2 c=0 \\ \text { and } & & 6 a-2 b+2 c=0 \\ \Rightarrow & & 2 a+3 b-c=0 \\ \text { and } & & 3 a-b+c=0 \end{array} $

For determining the point of concurrency with $a x+b y+c=0$, compare both equations with

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

∴ The lines are concurrent at $(-2,-3)$ and $(3,-1)$.

Given, $H \equiv a x^2-x y+b y^2=0$

$ \begin{aligned} \tan \theta & =\frac{2 \sqrt{h^2-a b}}{|a+b|} \\ \Rightarrow \quad 5 & =\frac{\sqrt[2]{\left(-\frac{1}{2}\right)^2-a b}}{|a+b|} \\ \Rightarrow \quad 5|a+b| & =2 \sqrt{\frac{1}{4}-a b} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Also, $(1,-1)$ is a point on $H=0$

$ a+1+b=0 \Rightarrow a=-(b+1) $

Substitute value of $a$ in Eq. (i),

$ \begin{aligned} & & 5|-1| & =2 \sqrt{\frac{1}{4}+(b+1) b} \\ \Rightarrow & & 25 & =4\left(\frac{1}{4}+b^2+b\right) \\ & \Rightarrow & 25 & =1+4 b^2+4 b \end{aligned} $

$ \begin{aligned} \Rightarrow & & 4 b^2+4 b-24 & =0 \\ \Rightarrow & & b^2+b-6 & =0 \\ \Rightarrow & & b^2+3 b-2 b-6 & =0 \\ \Rightarrow & & b & =2,-3 \end{aligned} $

If $b=2$, then $a=-3$

If $b=-3$, then $a=2$

$ a^2+a b+b^2=4-6+9=7 $

Given, line are

$ \begin{aligned} & 3 x^2-4 x y+5 y^2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & a=3, b=5, h=-2 \end{aligned} $

Line is passing $(2,3)$ and perpendicular to Eq. (i),

$ \begin{aligned} & \therefore 5(x-2)^2+4(x-2)(y-3)+3(y-3)^2=0 \\ & \Rightarrow 5 x^2+20-20 x+4(x y-3 x-2 y+6) \\ & \quad+3 y^2+27-18 y=0 \\ & \begin{aligned} \Rightarrow 5 x^2+4 x y+3 y^2-32 x-26 y+71=0 \\ \therefore a=5, b=3, c=71, f=-13, \\ \quad g=-16, h=2 \\ \therefore a+b+c+f+g+h \\ \quad=5+3+71-13-16+2 \\ \quad=8+71-27=8+44=52 \end{aligned} \end{aligned} $

Given, the equation of line $4 x-6 y-2=0$ and the equation of curve

$ 3 x^2-4 x y+5 y^2-2 x+y-6=0 $

The equation of line $4 x-6 y-2=0$ can be written as $2 x-3 y=1$

The equation of curve can be written as

$ \left(3 x^2-4 x y+5 y^2\right)-(2 x-y)(1)-6(1)^2=0 $

Now, substitute $1=2 x-3 y$ in equation of curve

$ \begin{aligned} & \left(3 x^2-4 x y+5 y^2\right)-(2 x-y)(2 x-3 y) -6(2 x-3 y)^2=0 \\ & \Rightarrow 3 x^2-4 x y+5 y^2-4 x^2+8 x y-3 y^2 \end{aligned} $

$ \begin{array}{rlrl} & -24 x^2-54 y^2+72 x y=0 \\ & - 25 x^2+76 x y-52 y^2=0 \\ & f(x, y) =25 x^2-76 x y+52 y^2 \\ f(1,-1) & =25+76+52=153 \\ & f(-1,-1) =25-76+52=1 \\ & \therefore \quad \frac{f(1,-1)}{f(-1,-1)} =153 \end{array} $

Given, $2 x-y=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

(a, b) lies on line $2 x-y-4=0$

$ \begin{aligned} b & =2 a-4 \\ \left(\frac{-4 m-n}{m+n}\right) & =2\left(\frac{m+2 n}{m+n}\right)-4 \\ -4 m-n & =2 m+4 n-4 m-4 n \\ 2 m+n & =0 \\or\,\,\,\,\,\,\,\, n & =-2 m \end{aligned} $

Equation of line joining the points $(2,-1)$ and $(1,-4)$ is

$ \begin{aligned} y+4 & =\left(\frac{-1+4}{2-1}\right)(x-1) \\ y+4 & =3(x-1) \\ y+4 & =3 x-3 \\ 3 x-y & =7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Subtracting Eq. (i) from Eq. (ii), we get

$ \begin{aligned} & x=3 \text { and } y=2 \\ & \therefore a=3, b=2 \\ & \text { So, } 4\left(a-b\left(\frac{m}{n}\right)^2\right)=4\left[3-2\left(\frac{-1}{2}\right)^2\right] \\ & \quad=4\left[3-\frac{1}{2}\right]=4 \times \frac{5}{2}=10 \end{aligned} $

Given, $x+(5 \lambda+1) y+1-3 \lambda=0$

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

The given family of lines contains two lines i.e. $x+y+1=0$

and    $ 5 y-3=0 $

$ \begin{gathered} 5 y-3=0 \Rightarrow y=\frac{3}{5} \\ x+\frac{3}{5}+1=0 \Rightarrow x=-\frac{8}{5} \end{gathered} $

The point of concurrent of family of lines

$ \begin{aligned} & x+(5 \lambda+1) y+1-3 \lambda \text { is }\left(\frac{-8}{5}, \frac{3}{5}\right) \\ & (5 \mu+2) x-3 y+3+6 \mu=0 \\ & 2 x-3 y+3+\mu(5 x+6)=0 \end{aligned} $

The family of lines contain

$ \begin{aligned} & 2 x-3 y+3=0 \\and\,\,\,\, & 5 x+6=0 \\ & x=\frac{-6}{5} \text { and } y=\frac{1}{5} \end{aligned} $

The point of concurrent of family of lines

$ (5 \mu+2) x-3 y+3+6 \mu=0 \text { is }\left(\frac{-6}{5}, \frac{1}{5}\right) $

∴ Required distance

$ \begin{aligned} & =\sqrt{\left(\frac{-8}{5}+\frac{6}{5}\right)^2+\left(\frac{3}{5}-\frac{1}{5}\right)^2} \\ & =\sqrt{\frac{4}{25}+\frac{4}{25}} \\ & =\sqrt{\frac{8}{25}}=\frac{2 \sqrt{2}}{5} \end{aligned} $

The two given lines $2 x-3 y+4=0$ and $6 x-9 y+7=0$ are parallel.

Slope of these lines $=\frac{2}{3}$ and slope of line $L=\frac{-3}{2}$

Equation of line $L$ is

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

Solving equation $3 x+2 y=5$ and $2 x-3 y+4=0$ to determine the point $A$.

$ x=\frac{7}{13}, y=\frac{22}{13} $

Solving equation $3 x+2 y=5$ and $6 x-9 y+7=0$ to get the point $B$.

$ x=\frac{31}{39} \text { and } y=\frac{17}{13} $

Coordinates of point $A$ is $\left(\frac{7}{13}, \frac{22}{13}\right)$ and point $B$ is $\left(\frac{31}{39}, \frac{17}{13}\right)$.

Let $P$ divides in the ratio $K: 1$

$ \begin{array}{cc} & 1=\frac{K \times \frac{31}{39}+1 \times \frac{7}{13}}{K+1} \\ \Rightarrow & 39(K+1)=31 K+21 \\ \Rightarrow & 39 K+39=31 K+21 \\ \Rightarrow & 8 K=-18 \\ & K=\frac{-9}{4} \end{array} $

$\therefore P$ divides $A B$ in the ratio $9: 4$ externally.

$ \text { Slope of line } B C=\frac{5+1}{-3-5} $

$ =\frac{6}{-8}=\frac{-3}{4} $

Slope of line $A D=\frac{4}{3}\,\,\,\,\, [\because A D \perp B C] $

Equation of $A D$ is

$ \begin{aligned} y-3 & =\frac{4}{3}(x-1) \\ 3 y-9 & =4 x-4 \\ 4 x-3 y & =-5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Slope of $A C=\frac{3+1}{1-5}=-1$

Slope of $B E=1$         $[B E \perp A C]$

Equation of line $B E$ is

$ \begin{aligned} y-5 & =1(x+3) \\ y & =x+3+5 \\ y & =x+8 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Substitute value of $y$ from Eq. (ii) to Eq. (i),

$ \begin{gathered} 4 x-3(x+8)=-5 \\ 4 x-3 x-24=-5 \\ x=19 \Rightarrow y=27 \end{gathered} $

∴ The coordinate of orthocentre is $(19,27)$.

$ \begin{aligned} & \text { Given, } \alpha x^2+2 \gamma x y+\beta y^2=0 \\ & \left(y-m_1 x\right)\left(y-m_2 x\right)=0 \\ & y^2-\left(m_1+m_2\right) x y+m_1 m_2 x^2=0 \\ & \Rightarrow \beta=1,2 r=-\left(m_1+m_2\right) \text { and } \alpha=m_1 m_2 \\ & \text { Again, } b h x^2+a b x y+a h y^2=0 \\ & \left(y+\frac{1}{m_1} x\right)\left(y+\frac{1}{m_2} x\right)=0 \end{aligned} $

$ \begin{array}{r} \left(m_1 y+x\right)\left(m_2 y+x\right)=0 \\ m_1 m_2 y^2+\left(m_1+m_2\right) x y+x^2=0 \end{array} $

Here, $a h=m_1 m_2, a b=m_1+m_2 a h=1$

Here, $a h=\beta, a b=-2 r$ and $\alpha=a h$

$ \frac{\alpha \beta}{r^2}=\frac{a h \times b h}{\left(\frac{a b}{-2}\right)^2}=\frac{4 h^2}{a b} $

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

Here, $h b=m^2, a b=2 m$ and $a h=1$

$ \begin{aligned} h b & =\left(\frac{a b}{2}\right)^2 \Rightarrow 4 h b=a^2 b^2 \\ 4 h & =a^2 b \Rightarrow 4 h \times h=a b \times a h \\ 4 h^2 & =a b \quad[\because a h=1] \end{aligned} $

Given, curve : $x^2+y^2-2(x+2 y)$

(1) $+2(1)^2=0$

Given, straight line $\frac{x+y}{k}=1$

Now, the line joining the origin to point of intersection of the given line and given curve is

$ \begin{array}{r} x^2+y^2-2(x+2 y)\left(\frac{x+y}{k}\right) +2\left(\frac{x+y}{k}\right)^2=0 \\ \Rightarrow k^2 x^2+k^2 y^2-2 k\left(x^2+3 x y+2 y^2\right) +2\left(x^2+y^2+2 x y\right)=0 \\ x^2\left(k^2-2 k+2\right)+y^2\left(k^2-4 k+2\right) +x y(-6 k+4)=0 \end{array} $

If the two homogenous lines are perpendicular, then coefficient of $x^2+$ Coefficient of $y^2=0$

$ \begin{aligned} k^2-2 k+2+k^2-4 k+2 & =0 \\ 2 k^2-6 k+4 & =0 \\ k^2-3 k+2 & =0 \\ k & =1,2 \end{aligned} $

∴ Sum of value of $k=1+2=3$

Let origin is shifted to $(p, q)$

$ \begin{aligned} & 3(X+p)^2+4(X+p)(Y+q)+(Y+q)^2 \\ & -8(X+p)-4(Y+q)-4=0 \\ & \Rightarrow 3 X^2+4 X Y+Y^2+(6 p+4 q-8) X \\ & +(4 p+2 q-4) Y+\left(3 p^2+4 p q+q^2\right. \\ & \quad-8 p-4 q-4)=0 \end{aligned} $

Now, $6 p+4 q-8=0,4 p+2 q-4=0$

$ \begin{array}{lrl} \Rightarrow & p & =0, q=2 \\ \therefore & f(X, Y) & =3 X^2+4 X Y+Y^2-8 \\ \Rightarrow & f(1,1) & =3+4+1-8=0 \end{array} $

$ \begin{aligned} &\text { According to the questions, }\\ &\begin{aligned} & \frac{m}{n}=-\frac{(2 \times(-2)-3 \times 3+4)}{(2 \times 3-3(-2)+4}=\frac{9}{16} \\ \therefore & \frac{-4 m}{3 n}=\frac{-4}{3} \times \frac{9}{16}=\frac{-3}{4} \end{aligned} \end{aligned} $

$ \begin{aligned} P & \equiv\left(\frac{3 \times 3-4 \times(-2)}{3-4}, \frac{3 \times(-2)-4(3)}{3-4}\right) \\ & =(-17,18) \end{aligned} $

$ \begin{aligned} &\text { According to question, }\\ &\left|\begin{array}{ccc} 2 & 1 & 3 \\ k & 2 & -3 \\ 3 & -2 & 1 \end{array}\right|=0 \end{aligned} $

$ \Rightarrow \quad 2(-4)-1(k+9)+3(-2 k-6)=0 $

$ \Rightarrow \quad-8-k-9-6 k-18=0 $

$ \begin{array}{rlrl} \Rightarrow &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, -7 k =35 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k =-5 \end{array} $

$ \therefore \quad L_2:-5 x+2 y-3=0 $

$ \begin{aligned} &\text { Let } \theta \text { be the angle between } L_2 \text { and }\\ &\begin{aligned} & 2 x-5 y+7=0 \\ & \therefore \tan \theta=\left|\frac{\frac{5}{2}-\frac{2}{5}}{1+\frac{5}{2} \times \frac{2}{5}}\right|=\frac{21}{20} \Rightarrow \cos \theta=\frac{20}{29} \end{aligned} \end{aligned} $

$ \text { According to the question, } $

$ \begin{aligned} &\begin{array}{rlrl} \frac{h-1}{1} & =\frac{k-1}{-1}=-2 \frac{(1-1+1)}{\left(1^2+1^2\right)}=-1 \\ & h =0, k=2 \\ ∴& Q =(0,2) \end{array}\\ &\text { ∴ Required length of perpendicular }\\ &=\frac{|3 \times 0-4 \times 2+3| \mid}{\sqrt{3^2+4^2}}=\frac{5}{5}=1 \end{aligned} $

$ \begin{aligned} & \text { Given, } \\ & x-3 y+3=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & x+3 y+3=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & x+y-1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

From Eqs. (i) and (ii), $x=-3, y=0$

From Eqs. (ii) and (iii), $x=3, y=-2$

From Eqs. (i) and (iii), $x=0, y=1$

∴ Vertices of given triangle are ( $-3,0$ ), ( $3,-2$ ) and $(0,1)$.

$ \begin{aligned} \therefore \text { Centroid } & =\left(\frac{-3+3+0}{3}, \frac{0-2+1}{3}\right) \\ & =\left(0, \frac{-1}{3}\right) \end{aligned} $

Given, equation of pair of lines

$ \begin{aligned} & 5 x^2+\frac{40}{3} x y+k y^2=0 \\ \Rightarrow & k\left(\frac{y}{x}\right)^2+\frac{40}{3}\left(\frac{y}{x}\right)+5=0 \end{aligned} $

Here, $y / x=3$ satisfies the equation

$ \therefore \quad 9 k+40+5=0 \Rightarrow k=-5 $

∴ Pair of straight line is

$ 5 x^2+\frac{40}{3} x y-5 y^2=0 $

As coefficients of $x^2$ and $y^2$ are negative of each other, so angle between the lines is right angle.

$\because 6 x^2-x y-12 y^2=0$

$ \begin{aligned} & \Rightarrow \quad(3 x+4 y)(2 x-3 y)=0 \\ & \text { and } \quad 15 x^2+14 x y-8 y^2=0 \\ & \Rightarrow \quad(3 x+4 y)(5 x-2 y)=0 \\ & \therefore \quad 3 x+4 y=0 \text { is common line } \end{aligned} $

∴ Combined equation of other two lines is

$ \begin{aligned} & (2 x-3 y)(5 x-2 y)=0 \\ \Rightarrow \quad & 10 x^2-19 x y+6 y^2=0 \end{aligned} $

Common line of two pairs is

$ 3 x+4 y=0 $

∴ Equation of $L$ is $(y-1)=\frac{-3}{4}(x+1)$

$ \Rightarrow L \text { is } 3 x+4 y-1=0 \Rightarrow 3 x+4 y=1 $

Homoginising given curve with

$ 3 x+4 y-1=0 $

we get

$ \begin{aligned} & 2 x^2-x y-y^2+(3 x+4 y) x -y(3 x+4 y)=0\\ & \Rightarrow 5 x^2-5 y^2=0 \Rightarrow x^2-y^2=0 \end{aligned} $

∴ Required equation of pair of straight lines is

$ x^2-y^2=0 $

Let $A(5,-3) B(3,-2) C(-1,5)$ be three points.

Let $P(h, k)$ satisfying

$ \begin{aligned} & (P A)^2+2(P B)^2=3(P C)^2 \\ & \begin{aligned} &(h-5)^2+(k+3)^2+2(h-3)^2+2(k+2)^2 \\ & \quad=3(h+1)^2+3(k-5)^2 \\ & \Rightarrow-10 h+25+6 k+9-12 h +18+8 k+8 \end{aligned} \end{aligned} $

$ \begin{aligned} & =6 h+3-30 k+75 \\ \Rightarrow \quad- & 28 h+44 k-18=0 \\ \Rightarrow \quad & 14 h-22 k+9=0 \end{aligned} $

Locus $14 x-22 y+9=0$

Only option (d), $\left(2, \frac{37}{22}\right)$ satisfies the locus.

Given, lines

$ \begin{aligned} & \frac{x}{a}+\frac{y}{b}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \frac{x}{b}+\frac{y}{a}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Slope of line (i), $m_1=\frac{-1 / a}{1 / b}=\frac{-b}{a}$

Slope of line (ii), $m_2=\frac{-1 / b}{1 / a}=\frac{-a}{b}$

Then, $\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$

$ =\left|\frac{-\frac{b}{a}+\frac{a}{b}}{1+1}\right|=\left|\frac{a^2-b^2}{2 a b}\right| $

Then, $\sin \theta=\left|\frac{a^2-b^2}{a^2+b^2}\right|$.

Let $A$ be the point of intersection of

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

$B$ is the point of intersection of

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

$ \begin{aligned} &\text { Sides of triangle are } x=2\\ &\begin{aligned} & 4 x+3 y+7=0 \\ & y=3 \\ & B C=a=\sqrt{6^2+8^2}=10 \\ & A C=b=6 \\ & A B=c=8 \end{aligned} \end{aligned} $

$ \begin{aligned} & \left(\frac{a x_1+b x_2+c x_2}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c}\right) \\ & I\left(\frac{(10 \times 2)+(6 \times 2)+(8 \times-4)}{10+6+8}\right. \\ & \left.\frac{(10 \times 3)+(6 \times-5)+(8 \times 3)}{10+6+8} I(0,1)\right) \end{aligned} $

As $\triangle A B C$ is right angled triangle.

So, mid-point of $B C$ is circumcentre

$ \begin{aligned} & S(-1,-1) \\ & I S=\sqrt{1^2+2^2}=\sqrt{5} \end{aligned} $

Equation $a x^2-4 x y-2 y^2=0$

represents pair of lines.

Also, $\cos \theta=\frac{1}{5} \Rightarrow \tan \theta=2 \sqrt{6}$

$ \begin{aligned} \because \quad \tan \theta & =\frac{2 \sqrt{h^2-a b}}{a+b} \\ 2 \sqrt{6} & =\frac{2 \sqrt{4+2 a}}{a-2} \end{aligned} $

Squaring both sides,

$ \begin{array}{rlrl} & & (a-2)^2 \cdot 6 & =4+2 a \\ \Rightarrow & & (a-2)^2 \cdot 3 & =2+a \\ \Rightarrow & & 3 a^2-12 a+12 & =2+a \\ \Rightarrow & & 3 a^2-13 a+10 & =0 \\ \Rightarrow & & (3 a-10)(a-1) & =0 \\ & & a=1, \frac{10}{3} \end{array} $

$ \begin{aligned} & \text { As } a_2>a_1 \quad \Rightarrow \quad a_1=1, a_2=\frac{10}{3} \\ & \therefore \quad a_1+3 a_2=11 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text {Lines represented by } \\ & 4 x^2-5 x y+3 y^2=0 \text { are } L_1 \text { and } L_2 \\ & 3\left(\frac{y}{x}\right)^2-5\left(\frac{y}{x}\right)+4=0 \end{aligned} \end{aligned} $

$ \begin{aligned} m_1+m_2 & =\frac{5}{3} \\ m_1 \cdot m_2 & =\frac{4}{3} \\ \frac{1}{m_1}+\frac{1}{m_2} & =\frac{m_1+m_2}{m_1 m_2}=\frac{5}{4} \end{aligned} $

Let the slope of line $L_1$ is $m_1$ and the slope of line $L_2$ is $m_2$.

$L_3$ is perpendicular to $L_1 \Rightarrow$ So, slope of

$ L_3=-\frac{1}{m_1} $

$L_4$ is perpendicular to $L_2 \Rightarrow$ So, slope of

$ L_4=-\frac{1}{m_2} $

Let the equation of $L_3: y=\frac{-1}{m_1} x+c_1$

It passes through $(4,3): 3=\frac{-4}{m_1}+c_1$

$ \begin{aligned} & \Rightarrow c_1=3+\frac{4}{m_1} \quad L_3: y=\frac{-1}{m_1} x+3+\frac{4}{m_1} \\ & \Rightarrow y+\frac{x}{m_1}-\left(3+\frac{4}{m_1}\right)=0 \end{aligned} $

and Equation of $L_4: y=-\frac{x}{m_2}+c_2$

Passes through $(4,3): 3=-\frac{4}{m_2}+c_2$

$ \begin{aligned} & \Rightarrow \quad c_2=3+\frac{4}{m_2} \\ & L_4: y=-\frac{x}{m_2}+3+\frac{4}{m_2} \\ & \Rightarrow \quad y+\frac{x}{m_2}-\left(3+\frac{4}{m_2}\right)=0 \end{aligned} $

Combined equation of $L_3$ and $L_4$

$ \left[y+\frac{x}{m_1}-\left(3+\frac{4}{m_1}\right)\right] $

$ \begin{aligned} & {\left[y+\frac{x}{m_2}-\left(3+\frac{4}{m_2}\right)\right]=0 } \\ & \Rightarrow y^2+\frac{x y}{m_2}-y\left(3+\frac{4}{m_2}\right)+\frac{x y}{m_1}+\frac{x^2}{m_1 m_2} \\ &- \frac{x}{m_1}\left(3+\frac{4}{m_2}\right)-\frac{x}{m_2}\left(3+\frac{4}{m_1}\right) \\ &-y\left(3+\frac{4}{m_1}\right)+\left(3+\frac{4}{m_1}\right)\left(3+\frac{4}{m_2}\right)=0 \\ & \Rightarrow y^2+\frac{x^2}{m_1 m_2}+x y\left(\frac{1}{m_1}+\frac{1}{m_2}\right) \end{aligned} $

$ \begin{array}{r} -x\left(\frac{8}{m_1 m_2}+3\left(\frac{1}{m_1}+\frac{1}{m_2}\right)\right) \\ -y\left(6+4\left(\frac{1}{m_1}+\frac{1}{m_2}\right)\right) \\ +\left(9+12\left(\frac{1}{m_1}+\frac{1}{m_2}\right)+\frac{16}{m_1 m_2}\right)=0 \\ \Rightarrow y^2+\frac{5}{4} x y+\frac{3}{4} x^2-x\left(6+3 \times \frac{5}{4}\right) \\ -y\left(6+4 \times \frac{5}{4}\right) \end{array} $

$ \begin{array}{r} +\left(9+12 \times \frac{5}{4}+16 \times \frac{3}{4}\right)=0 \\ \Rightarrow y^2+\frac{5}{4} x y+\frac{3}{4} x^2-\frac{39}{4} x-11 y+36=0 \\ \Rightarrow 4 y^2+5 x y+3 x^2-39 x-44 y+144=0 \end{array} $

$ \begin{aligned} &\text { On comparing with }\\ &\begin{aligned} & a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 \\ & a=4, h=\frac{5}{2}, b=3, g=\frac{-39}{2}, f=-22 \text { and } \\ & c=144 \\ & a f+b g+c h=-88-\frac{117}{2}+360=213.5 \end{aligned} \end{aligned} $

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

On comparing with

$ \begin{aligned} & a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 \\ & a=1, h=0, b=-1, g=\frac{a}{2}, f=0 \text { and } \\ & c=b \end{aligned} $

If it represents a pair of straight line

Then, $D=0$

$ \begin{aligned} & \Rightarrow p b c+2 f g h-a f^2-b g^2-c h^2=0 \\ & -b+0-0+\frac{a^2}{4}-0=0 \Rightarrow a^2=4 b \end{aligned} $

From the options, $a=6$ and $b=9$ satisfy above relation.

So, pair of $(a, b) \equiv(6,9)$

If coordinate axes are rotated through an angle $\theta$ in anti-clockwise direction, then

$ x=X \cos \theta-Y \sin \theta $

and $y=X \sin \theta+Y \cos \theta$, on substituting the values of $x$ and $y$ in equation $x^2+y^2+2 x y+2 x+6 y+1=0$, we get $(X \cos \theta-Y \sin \theta)^2+(X \sin \theta+Y \cos \theta)^2$

$ \begin{array}{r} +2(X \cos \theta-Y \sin \theta)(X \sin \theta+Y \cos \theta) \\ +2(X \cos \theta-Y \sin \theta) \\ +6(X \sin \theta+Y \cos \theta)+1=0 \end{array} $

$ \begin{aligned} & \Rightarrow\left(\cos ^2 \theta+\right.\left.\sin ^2 \theta+2 \sin \theta \cos \theta\right) X^2 \\ &+(-2 \sin \theta \cos \theta+2 \sin \theta \cos \theta \\ &\left.+2 \cos ^2 \theta-2 \sin ^2 \theta\right) X Y \\ &+\left(\sin ^2 \theta+\cos ^2 \theta-2 \sin \theta \cos \theta\right) Y^2 \end{aligned} $

$ \begin{array}{r} +(2 \cos \theta+6 \sin \theta) X+(6 \cos \theta-2 \sin \theta) Y \\ +1=0 \\ \Rightarrow(1+\sin 2 \theta)+2 \cos 2 \theta X Y+(1-\sin 2 \theta) Y^2 \\ +(2 \cos \theta+6 \sin \theta) X \\ +(6 \cos \theta-2 \sin \theta) Y+1=0 \end{array} $

On comparing with transformed given equation

$ \begin{aligned} (2+\sqrt{3}) X^2+2 X Y & +(2-\sqrt{3}) Y^2 \\ & +a X+b Y+2=0, \end{aligned} $

we get $\sin 2 \theta=\frac{\sqrt{3}}{2}, \cos 2 \theta=\frac{1}{2}$

$ \begin{aligned} & \Rightarrow \quad \theta=\frac{\pi}{6} \\ & \therefore \quad a=4(\cos \theta+3 \sin \theta)=2 \sqrt{3}+6 \\ & \text { and } b=4(3 \cos \theta-\sin \theta)=6 \sqrt{3}-2 \\ & \therefore 3 a-b=6 \sqrt{3}+18-6 \sqrt{3}+2=20 \end{aligned} $

$ \text { Let rectangle be } A B C D \text {. } $

Here line $A B \| C D$

⇒ slope of line $A B=$ slope of line $C D$

$ \begin{array}{ll} \Rightarrow & -3=\frac{-b}{2} \\ \Rightarrow & b=6 \end{array} $

Now line $A B \perp^r B C$

⇒ (slope of line $A B) \times$ (slope of line

$ \begin{gathered} B C)=-1 \\ \Rightarrow \quad(-3)\left(\frac{1}{a}\right)=-1 \Rightarrow a=3 \\ \therefore \text { Eq. of } B C: x-3 y-10=0 \end{gathered} $

Eq. of $C D: 6 x+2 y+9=0$

∴ Distance between $A B$ and $C D$

$ \begin{aligned} & A B: 3 x+y-4=0 \\ \Rightarrow & C D: 3 x+y+\frac{9}{2}=0 \\ \therefore & \text { i.e., } B C=\left|\frac{-4-9 / 2}{\sqrt{9+1}}\right|=\frac{17}{2 \sqrt{10}} \end{aligned} $

Now line $A D$ pass through the point $(1,2)$

⇒ Distance $C D=$ distance of point $(1,2)$ from line $B C$

$ =\left|\frac{1-6-10}{\sqrt{1+9}}\right|=\frac{15}{\sqrt{10}} $

Now, area of rectangle $A B C D=B C \times C D$

$ =\frac{17}{2 \sqrt{10}} \times \frac{15}{\sqrt{0}}=\frac{17 \times 15}{2 \times 10}=\frac{51}{4} $

$ \text { Given, } A B C D \text { is a rectangle } $

Slope of $A B=\frac{-2-1}{3-2}=-3=m_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Slope of $B C=\frac{b+2}{a-3}=m_2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now $A B \perp B C \Rightarrow m_1 m_2=-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

∴ Putting value of $m_1$ and $m_2$ in Eq. (iii)

$ \begin{align*} -3\left(\frac{b+2}{a-3}\right) & =-1 \\ -3 b-6 & =-a+3 \\ \Rightarrow a-3 b-9 & =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) \end{align*} $

Now, slope of line $C D=$ slope of $C P$

$ \begin{gathered} m_3=\frac{b-4}{a-3} \\ \therefore \text { line } A B \| C D \Rightarrow m_1=m_3 \end{gathered} $

$ \begin{array}{rlrl} \therefore & -3 =\frac{b-4}{a-3} \\ -3 a+9 & =b-4 \end{array} $

$ \begin{equation*} \Rightarrow \quad 3 a+b-13=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v) \end{equation*} $

Now, solving Eqs. (iv) and (v),

$ a-3 b-9=0 \Rightarrow 3 a+b-13=0 $

Eq. (iv) $\times(3)$ and subtracting Eq. (iv) and Eq. (v),

$ \begin{aligned} &\Rightarrow \quad b=\frac{-14}{10} \Rightarrow b=\frac{-7}{5}\\ &\text { These value is putting in Eq. (iv) }\\ &\begin{aligned} & & a-3\left(-\frac{7}{5}\right)-9 & =0 \\ \Rightarrow & & a+\frac{21}{5}-9 & =0 \end{aligned} \end{aligned} $

$ \begin{aligned} &a+\left(\frac{24}{5}\right)=0 \Rightarrow a=\frac{24}{5}\\ &\text { Now, } \begin{aligned} 5 a+10 b & =5 \times \frac{24}{5}+10 \times\left(-\frac{7}{5}\right) \\ & =24-14=10 \end{aligned} \end{aligned} $

Given, $\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=0$ ⇒ lines may be concurrent or two pair of lines be parallel Now, if $\frac{a_i}{a_j} \neq \frac{b_i}{b_j} \neq \frac{c_i}{c_j}(i \neq j)$ then lines are concurrent.

Equation of given pair of straight lines

$S=3 x^2-2 k x y+y^2=0$ and the line

$ L=2 x-y-6=0 $

On solving, we get

$ \begin{aligned} 3 x^2-2 k x(2 x-6)+(2 x-6)^2 & =0 \\ \Rightarrow \quad(7-4 k) x^2+12(k-2) x+36 & =0 \end{aligned} $

Let the intersection points are $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$

∴ The area of $\triangle A B C$ is

$ \frac{1}{2}\left|\left|\begin{array}{ccc} 0 & 0 & 0 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{array}\right|\right|=36 \Rightarrow\left|\frac{y_1}{x_1}-\frac{y_2}{x_2}\right|=\frac{72}{\left|x_2 x_1\right|} $

$ \begin{aligned} & \because m_1+m_2=2 k, m_1 m_2=3 \text { and } \frac{y_1}{x_1}=m_1 \\ & \qquad \frac{y_2}{x_2}=m_2 \text { and }\left|x_1 x_2\right|=\frac{36}{|7-4 k|} \\ & \text { So, }\left|m_1-m_2\right|=\frac{72}{36}|7-4 k| \\ & \Rightarrow \quad\left(m_1+m_2\right)^2-4 m_1 m_2=4(7-4 k)^2 \\ & \quad(\text { on squaring both sides }) \\ & \Rightarrow \quad 4 k^2-12=4\left(49+16 k^2-56 k\right) \\ & \Rightarrow \quad 15 k^2-56 k+52=0 \\ & \Rightarrow \quad 15 k^2-30 k-26 k+52=0 \\ & \Rightarrow \quad 15 k(k-2)-26(k-2)=0 \\ & \Rightarrow \quad \quad k=2 \quad(\because k \text { is an integer }) \\ & \therefore \quad \tan \theta=\frac{2 \sqrt{k^2-3}}{3+1}=\frac{1}{2} \\ & \Rightarrow \quad \sin \theta=\frac{1}{\sqrt{5}} \end{aligned} $

For $\triangle A B C$, we have sides are

$ \begin{array}{r} 2 x^2-y^2=0, x+y-1=0 \\ (\sqrt{2} x-y)(\sqrt{2} x+y)=0, x+y-1=0 \\ \sqrt{2} x-y=0, \sqrt{2} x+y=0, x+y-1=0 \end{array} $

On solving these, we get vertices

$ (0,0),(\sqrt{2}-1,2-\sqrt{2}),(-\sqrt{2}-1,2+\sqrt{2}) $

For $\triangle P Q R$, we have sides are

$ \begin{array}{r} 2 x^2-5 x y+2 y^2=0,7 x-2 y-12=0 \\ (2 x-y)(x-2 y)=0,7 x-2 y-12=0 \\ 2 x-y=0, x-2 y=0,7 x-2 y-12=0 \end{array} $

On solving these, we get vertices

$ (0,0),(4,8),(2,1) $

Now, centroid of $\triangle A B C$

$ \begin{aligned} G & =\left(\frac{0+\sqrt{2}-1-\sqrt{2}-1}{3}, \frac{0+2-\sqrt{2}+2+\sqrt{2}}{3}\right) \\ & =\left(\frac{-2}{3}, \frac{4}{3}\right) \end{aligned} $

Orthocentre of $\triangle P Q R$,

$ \begin{aligned} H & =\left(\frac{(8-1)(8+8)}{16-4}, \frac{(2-4)(8+8)}{16-4}\right) \\ & =\left(\frac{28}{3},-\frac{8}{3}\right) \end{aligned} $

∴ Required distance $=G H$

$ \begin{aligned} & =\sqrt{\left(\frac{28}{3}+\frac{2}{3}\right)^2+\left(\frac{-8}{3}-\frac{4}{3}\right)^2} \\ & =\sqrt{100+16}=\sqrt{116}=2 \sqrt{29} \end{aligned} $

Given,

$ A=(2,3), B(3,-5), C=(x, y) $

Let centroid of $\triangle A B C$ is $(h, k)$

$ \begin{array}{ll} \therefore & h=\frac{2+3+x}{3} \\ & k=\frac{3-5+y}{3} \\ \Rightarrow & x=3 h-5 \\ \Rightarrow & y=3 k+2 \end{array} $

$C$ lie on line $3 x+4 y-5=0$

$ \begin{aligned} & \therefore 3(3 h-5)+4(3 k+2)-5=0 \\ & \Rightarrow \quad 9 h-15+12 k+8-5=0 \\ & \Rightarrow \quad 9 h+12 k-12=0 \\ & \Rightarrow \quad 3 h+4 k-4=0 \end{aligned} $

Locus of centroid is

$ 3 x+4 y-4=0 $

Which is parallel to line $L=0$

$ \begin{aligned} &\\ &\begin{aligned} & \text { Given, } 4 x+3 y+2=0 \\ & \quad \begin{array}{r}\Rightarrow -4 x-3 y=2 \\ \frac{-4 x}{5}-\frac{3}{5} y=\frac{2}{5} \end{array} \\ & \because \cos \alpha=\frac{-4}{5}, \sin \alpha=\frac{-3}{5} \\ & \text { and } \quad P=\frac{2}{5} \\ & \frac{x}{2 /-4}+\frac{y}{2 /-3}=1 \\ & \Rightarrow \quad a=\frac{-2}{4}, b=\frac{-2}{3} \\ & \therefore \frac{P \sec \alpha}{a b}=\frac{\frac{2}{5} \times\left(\frac{-5}{4}\right)}{\left(\frac{-2}{4}\right)\left(\frac{-2}{3}\right)}=\frac{-3}{2} \end{aligned} \end{aligned} $

Given, lines $x+2 y-5=0$ and $3 x-y+1=0$

$ \begin{aligned} &\text { lie in the same region of lines }\\ &\begin{aligned} & \begin{array}{l} x+2 y-5=0 \text { and } 3 x-y+1=0 \\ \left(k^2+2(k+1)-5\right)<0 \end{array} \\ & \text { and } \quad 3 k^2-(2 k+1)+1>0 \\ & \quad \begin{array}{r} \Rightarrow\,\, k^2+2 k-3<0 \\ \text { and } \quad 3 k^2-k>0 \end{array} \\ & \text { and } \quad \begin{aligned} &(k+3)(k-1)<0 \\ & k(3 k-1)>0 \\ & k \in(-3,1) \end{aligned} \\ & \text { and } \quad k \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right) \\ & \therefore \quad k \in(-3,0) \cup\left(\frac{1}{3}, 1\right) \end{aligned} \end{aligned} $

Equation of line are $x+y-1=0$ and $x-y+1=0$

Solving equation we get, $(0,1)$

$ \therefore R(0,1) $

Perpendicular distance from $P$ to line $x+y-1=0$ and $x-y+1=0$ are $\frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{2}}$ respectively

$ \therefore \quad P S=P Q $

and $\triangle P S R$ and $\triangle P Q R$ are congruent triangle

∴ Area of quadrilateral $P Q R S=P Q \times P S$

$ =\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{1}{2} $

Equation of curves are,

$ x^2+y^2+g x+c=0 $

and $x^2+y^2+2 f y-c=0$

Subtract Eq. (i) from Eq. (ii), we get

$ 2 f y-g x=2 c \Rightarrow \frac{2 f y-g x}{2 c}=1 $

Homogeneousing the Eq. (i), we get

$ \begin{aligned} & x^2+y^2+g x \frac{(2 f y-g x)}{2 c}+\frac{C(2 f y-g x)^2}{4 c^2}=0 \\ & \Rightarrow 4 c\left(x^2+y^2\right)+4 g f x y-2 g^2 x^2+4 f^2 y^2 +g^2 x^2-4 g f x y=0 \\ & \Rightarrow \quad\left(4 c-2 g^2+g^2\right) x^2+\left(4 c+4 f^2\right) y^2=0 \end{aligned} $

∴ A pair of straight formed right angle, then

$ \begin{array}{r} 4 c-g^2+4 f^2+4 c=0 \\ g^2-4 f^2=8 c \end{array} $

We have, pair of straight line is

$ \begin{array}{r} x^2-y^2-x+3 y-2=0 \\ (x-y+1)(x+y-2)=0 \\ x-y+1=0 \text { and } x+y-2=0 \end{array} $

Solving these equation, we get

$ x=\frac{1}{2} \text { and } y=\frac{3}{2} $

∴ Intersection point $P\left(\frac{1}{2}, \frac{3}{2}\right)$

$ \alpha=O P=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{3}{2}\right)^2}=\frac{\sqrt{10}}{2} $

Perpendicular distance from origin to line $x-y+1=0$

and $x+y-2=0$ are $\frac{1}{\sqrt{2}}$ and $\frac{2}{\sqrt{2}}$ respectively,

$ \begin{aligned} &\begin{aligned} \beta & =\frac{1}{\sqrt{2}} \times \frac{2}{\sqrt{2}}=1 \\ \Rightarrow \quad \alpha \beta & =\frac{\sqrt{10}}{2} \times 1=\sqrt{\frac{5}{2}} \end{aligned}\end{aligned} $

Given equation of line $x-y=0$ Distance from $P(\alpha, \beta)$ to line $x-y=0$ is

$ b=\left|\frac{\alpha-\beta}{\sqrt{2}}\right| \Rightarrow(\alpha-\beta)^2=2 b^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Distance between $P(\alpha, \beta)$ and $A(2,1)$ is

$ a^2=(\alpha-2)^2+(\beta-1)^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Distance from origin to $A(2,1)$ is $C^2=5$

Also $\quad a=b c$

$ \therefore \quad a^2=5 b^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) $

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

$ \begin{aligned} & (\alpha-2)^2+(\beta-1)^2=\frac{5(\alpha-\beta)^2}{2} \\ & \begin{aligned} 2\left(\alpha^2-4 \alpha+4+\beta^2-2 \beta+1\right] & \\ =5\left[\alpha^2+\beta^2-2 \alpha \beta\right] & \\ \Rightarrow 3\left(\alpha^2+\beta^2\right)-10 \alpha \beta+8 \alpha+4 \beta & -10=0 \end{aligned} \end{aligned} $

∴ Locus of $P$ is

$ 3\left(x^2+y^2\right)-10 x y+8 x+4 y-10=0 $

We have,

The point $(4,1)$ undergoes the following transformations

(i) Reflection in the line $x-y=0$

∵ Reflection of point $(4,1)$ to line $x-y=0$ is $(1,4)$

(ii) Shifting through a distance of 2 units along the positive $X$-axis then $(1+2,4) \equiv(3,4)$

(iii) Projection of $(3,4)$ in $X$-axis is $(3,0)$

∴ The new coordinates is $(3,0)$