Straight Lines and Pair of Straight Lines

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} $