Straight Lines and Pair of Straight Lines

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 $(3,0)$

Given line

$ 3 x+4 y=5 \text { and } $

$4 x-3 y=15$

Slope of line $3 x+4 y=5$ is

i.e.,        $ m_1=\frac{-3}{4} $

Slope of line $4 x-3 y=15$

i.e.,        $ m_2=\frac{4}{3} $

Let the slope of line $B C=m$

Given,$\,\,\,\,A B=A C$

$\because\,\,\,\,\tan \alpha=\tan \beta$

$ \begin{aligned} \frac{m-\frac{4}{3}}{1+\frac{4 m}{3}} & =\frac{-\frac{3}{4}-m}{1-\frac{3 m}{4}} \\ \frac{3 m-4}{3+4 m} & =\frac{-3-4 m}{4-3 m} \\ 12 m^2-9 m^2-16+12 m & \\ =-9-12 m-12 m-16 m^2 & \\ \Rightarrow \quad 7 m^2+48 m-7 & =0 \\ (7 m-1)(m+7) & =0 \\ m=1 / 7, m & =-7 \end{aligned} $

Equation of line passing through $(1,2)$ and slope $\frac{1}{7}$ is,

$ =y-2=\frac{1}{7}(x-1)=x-7 y+13=0 $

Equation of line passing through $(1,2)$ and slope $=-7$ is

$ \begin{aligned} y-2 & =-7(x-1) \\ 7 x+y & =9 \end{aligned} $

We know that the equation of two straight lines which passes through a point ( $x_1, y_1$ ) and made a given angle $\alpha$ with the given straight line

$ \begin{aligned} & y=m x+c \text { are } \\ & y-y_1=\frac{m \pm \tan \alpha}{1 \mp m \tan \alpha}\left(x-x_1\right) \end{aligned} $

Here, $x_1=1, y_1=1, \alpha=60^{\circ}$ and $m=-1$

(slope of line $x+y-3=0$ )

So, equation of lines are

$ y-1=\frac{-1-\tan 60^{\circ}}{1+(-1) \tan 60^{\circ}}(x-1) $

and $y-1=\frac{-1+\tan 60^{\circ}}{1-(-1) \tan 60^{\circ}}(x-1)$

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

and $\quad(1-\sqrt{3}) x+(1+\sqrt{3}) y-1-\sqrt{3}$

$ \begin{array}{lr} & -1+\sqrt{3}=0 \\ \Rightarrow & (1+\sqrt{3}) x+(1-\sqrt{3}) y-2=0 \\ \text { and } & (1+\sqrt{3}) x+(1-\sqrt{3}) y-2=0 \end{array} $

Given pairs of lines

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

∴ Pairs of line are

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

Solving these equation we get,

$ \begin{array}{ll} & x=1, y=2 \\ \therefore & \alpha=1 \end{array} $

Equation of line passing through origin is

$ y-2 x=0 $

∵ Slope of line $m=2$

$ \therefore \quad \beta=2 $

Equation of angular bisector of line

$ \begin{aligned} & \quad \frac{2 y-x-3}{\sqrt{5}}= \pm \frac{y-2 x}{\sqrt{5}} \\ & 2 y-x-3=y-2 x \text { or } \\ & 2 y-x-3=-y+2 x \\ & \because \quad x+y-3=0 \text { or } 3 x-3 y+3=0 \\ & \quad \quad x+y-3=0 \text { or } x-y+1=0 \\ & \quad \gamma=-3 \\ & \therefore \gamma<\alpha<\beta \text { i.e., }-3<1<2 \end{aligned} $

$ \begin{aligned} & \left(7 x^2-4 x y+8 y^2\right)^2+(4 x-8 y-32) \\ & \left(7 x^2-4 x y+8 y^2\right)=0 \end{aligned} $

$ \begin{aligned} \Rightarrow & \left(7 x^2-4 x y+8 y^2\right)\left(7 x^2-4 x y+8 y^2\right. +4 x-8 y-32) =0\end{aligned} $

Now, $7 x^2-4 x y+8 y^2+4 x-8 y-32=0$

$ \Rightarrow 7 x^2+(-4 y+4) x+8 y^2-8 y-32=0 $

Which is quadratic equation in $x$

$ \begin{aligned} & \therefore x=\frac{(4 y-4) \pm \sqrt{\begin{array}{c} \left(16 y^2+16-32 y\right)-4 \times 7 \\ \times\left(8 y^2-8 y-32\right) \end{array}}}{2 \times 7} \\ & \Rightarrow \\ & x=\frac{(2 y-2) \pm \sqrt{\begin{array}{c} 4 y^2+4-32 y \\ -56 y^2+56 y+224 \end{array}}}{7} \\ & \Rightarrow x=\frac{(2 y-2) \pm \sqrt{-52 y^2+24 y+228}}{7} \\ & \quad=\frac{(2 y-2) \pm \sqrt{-4\left(13 y^2-6 y-57\right)}}{7} \end{aligned} $

Here, $D<0$

So, lines are not possible.

Therefore, question is wrong.

We have, $R(h, k)$ is mid-point of $O M$.

$ \begin{aligned} & \therefore \quad\left(\frac{0+\alpha}{2}, \frac{0+\beta}{2}\right)=(h, k) \\ & \Rightarrow \alpha=2 h, \beta=2 k \end{aligned} $

⇒ Coordinates of $M$ are $(2 h, 2 k)$.

Now, slope of $O M=\frac{2 k-0}{2 h-0}=\frac{k}{h}$ and slope of $M Q=\frac{2 k-b}{2 h-a}$

Since, $O M \perp M Q$

$ \begin{aligned} & \therefore \quad \frac{k}{h} \times \frac{2 k-b}{2 h-a}=-1 \\ & \Rightarrow 2 k^2-b k=-2 h^2+a h \\ & \Rightarrow \quad 2 h^2+2 k^2-a h-b k=0 \end{aligned} $

∴ Locus of $R(h, k)$ is

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

Substituting $x=X+\frac{3}{2}, y=Y+\frac{3}{2}$ in the equation

$ 32 x^2+8 x y+32 y^2-108 x-108 y+99=0 $

we get

$ \begin{aligned} & 32\left(X+\frac{3}{2}\right)^2+8\left(X+\frac{3}{2}\right)\left(Y+\frac{3}{2}\right) +32\left(Y+\frac{3}{2}\right)^2-108\left(X+\frac{3}{2}\right) -108\left(Y+\frac{3}{2}\right)+99=0 \\ & \Rightarrow 32\left[X^2+3 X+\frac{9}{4}\right] +8\left(\frac{2 X+3}{2}\right)\left(\frac{2 Y+3}{2}\right)+32\left[Y^2+3 Y+\frac{9}{4}\right] \\ & \Rightarrow 32 X^2+96 X+72+2(4 X Y+6 X +6 Y+9)+32 Y^2 \end{aligned} $

$ \begin{aligned} +96 Y+72-108 X-162-108 Y \\ -162+99=0 \\ \Rightarrow 32 X^2+32 Y^2+8 X Y-63=0 \end{aligned} $

Equation of $L_1$ :

$ \begin{aligned} & y-4=1(x-3) \\ & y-4=x-3 \\ & \quad x-y+1=0 \end{aligned} $

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

Since $A C=B C$

$\therefore \triangle A B C$ is isosceles

$ \begin{aligned} \therefore \quad \angle C & =\angle B \\ \left|\frac{m-2}{1+2 m}\right|=\left|\frac{2-1}{1+2 \times 1}\right| & \\ & {\left[\because \text { slope of } B C=\frac{-2}{-1}=2\right] } \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{m-2}{1+2 m}= \pm \frac{1}{3} \\ & \Rightarrow \quad \frac{m-2}{1+2 m}=\frac{1}{3} \text { or } \frac{m-2}{1+2 m}=\frac{-1}{3} \\ & \Rightarrow \quad m=7 \text { or } m=1 \end{aligned} $

But for $m=1$

$A C$ become $L_1$

$ \therefore \quad m=7 $

Equation of $A C$ is

$ \begin{gathered} y-4=7(x-3) \\ y-4=7 x-21 \\ 7 x-y-17=0 \end{gathered} $

Let slope of line is $m$

∴ Equation of line is

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

On solving $m x-y+(2-m)$ and $x+y=4$, we get

$ (x, y)=\left(\frac{m+2}{m+1}, \frac{3 m+2}{m+1}\right) $

Now, it is given that

$ \begin{aligned} & \sqrt{\left(\frac{m+2}{m+1}-1\right)^2+\left(\frac{3 m+2}{m+1}-2\right)^2}=\frac{\sqrt{6}}{3} \\ & \Rightarrow \quad \frac{1}{(m+1)^2}+\frac{m^2}{(m+1)^2}=\frac{6}{9} \\ & \Rightarrow \quad \frac{1+m^2}{(m+1)^2}=\frac{2}{3} \\ & \Rightarrow \quad 3+3 m^2=2 m^2+4 m+2 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad m^2-4 m+1=0 \Rightarrow m=2 \pm \sqrt{3} \\ & \Rightarrow m=2+\sqrt{3}, 2-\sqrt{3} \end{aligned}\\ &\text { ∴ Angles will be } \frac{\pi}{12} \text { and } \frac{5 \pi}{12} \text {. } \end{aligned} $

It is given that lines $x+3 y-9=0$, $4 x+5 y-1=0, p x+q y+10=0$ are concurrent

$ \begin{aligned} & \therefore\left|\begin{array}{lll} 1 & 3 & -9 \\ 4 & 5 & -1 \\ p & q & 10 \end{array}\right|=0 \\ & \Rightarrow 50+q-3(40+p)-9(4 q-5 p)=0 \\ & \Rightarrow 50+q-120-3 p-36 q+45 p=0 \\ & \Rightarrow 42 p-35 q-70=0 \\ & \Rightarrow-6 p+5 q+10=0 \end{aligned} $

So, $5 x+6 y+10=0$ passes through $(q,-p)$.

The equation $(2 l-3) x^2+2 l x y-y^2=0$ will represent a pair of real and distinct lines, if

$ \begin{array}{ll} & l^2-(2 l-3)(-1)>0 \quad\left[\because h^2>a b\right] \\ \Rightarrow \quad & l^2+2 l-3>0 \\ \Rightarrow \quad & (l+3)(l-1)>0 \\ \therefore \quad & l \in \mathbf{R}-(-3,1) \end{array} $

We have,

$ \begin{array}{ll} & 2 y^2-x y-6 x^2=0 \\ \Rightarrow & 2 y^2-4 x y+3 x y-6 x^2=0 \\ \Rightarrow & 2 y(y-2 x)+3 x(y-2 x)=0 \\ \Rightarrow & (y-2 x)(2 y+3 x)=0 \end{array} $

∴ Equations of sides of triangle are

$ x+y=1, y-2 x=0,2 y+3 x=0 $

On solving the above equations, we get the vertices of triangle as $\left(\frac{1}{3}, \frac{2}{3}\right),(0,0)$ and $(-2,3)$.

∴ Centroid of triangle

$ \begin{aligned} & =\left(\frac{\frac{1}{3}+0-2}{3}, \frac{\frac{2}{3}+0+3}{3}\right) \\ & =\left(\frac{-5}{9}, \frac{11}{9}\right) \end{aligned} $

A line passing through (1, 2) and parallel to the line 3x $-$ y = 10 is

3(x $-$ 1) $-$ (y $-$ 2) = 0

$\Rightarrow$ 3x $-$ 3 $-$ y + 2 = 0 $\Rightarrow$ 3x $-$ y $-$ 1 = 0

The intersection point of line

x + y + 5 = 0 ..... (i)

and 3x $-$ y $-$ 1 = 0 ..... (ii)

are ($-$1, $-$4)

$\therefore$ Distance from the line x + y + 5 to the point (1, 2) measured parallel to the line 3x $-$ y = 10 is

$\sqrt {{{( - 1 - 1)}^2} + {{( - 4 - 2)}^2}} $

$\sqrt {4 + 36} = \sqrt {40} = 2\sqrt {10} $

For three lengths to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. This is known as the triangle inequality theorem.

Let's apply this to our lengths: $5$, $5r$, and $5r^2$. We have three inequalities to check:

1. $5 + 5r > 5r^2$
2. $5 + 5r^2 > 5r$
3. $5r + 5r^2 > 5$

Solving these inequalities will give us a range of values for $r$. Any value outside of this range will mean that $r$ cannot form a triangle with the given sides.

1. 1. $1 + r > r^2$

We can rewrite this as: $r^2 - r - 1 < 0$

$\Rightarrow\left[r-\left(\frac{1-\sqrt{5}}{2}\right)\right]\left[r-\left(\frac{1+\sqrt{5}}{2}\right)\right]<0$

The roots of this quadratic equation are $\frac{1-\sqrt{5}}{2}$ and $\frac{1+\sqrt{5}}{2}$, so the inequality holds for $r$ in the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$.

1. 2. $1 + r^2 > r$

We can rewrite this as: $r^2 - r + 1 > 0$

This inequality is true for all real numbers since the left-hand side is always positive, which means the inequality holds for all real $R$.

1. 3. $r + r^2 > 1$

We can rewrite this as: $r^2 + r - 1 > 0$

Taking the intersection of the intervals from the three inequalities, we get $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ as the valid range for $r$.

Given the solution of the inequalities $r \in \left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$, we need to determine which of the options are outside this interval.

1. Option A: $\frac{7}{4}$

The decimal representation of $\frac{7}{4}$ is 1.75, which is outside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ or approximately (-0.618, 1.618). Hence, r cannot be equal to $\frac{7}{4}$.

1. Option B: $\frac{5}{4}$

The decimal representation of $\frac{5}{4}$ is 1.25, which is inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{5}{4}$.

1. Option C: $\frac{3}{4}$

The decimal representation of $\frac{3}{4}$ is 0.75, which is also inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{3}{4}$.

1. Option D: $\frac{3}{2}$

The decimal representation of $\frac{3}{2}$ is 1.5, which is also inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{3}{2}$.

Therefore, the only value that r cannot be equal to, among the given options, is $\frac{7}{4}$. So, the correct answer is:

Option A: $\frac{7}{4}$