Question

The straight line $L=x+y+1=0$and $L_{1}x+2y+3=0$ are intersecting. $m$ is the slope of the straight line $L_2$ such that $L$ is the bisector of the angle between $L_1$ and $L_2$. The value of $812m^2+3$ is equal to

Answer 1

Let the equation of $L_2$ be $L_1+\lambda L=0$
$\Rightarrow (1+\lambda )x+(2+\lambda )y+3+\lambda =0$
slopes of $L_2,L$ and $L_1$ are
$-\frac{1+\lambda }{2+\lambda },-1,-1/2$
Since $L$ is the bisector of the angle between $L_1$ & $L_2$
$\therefore \frac{-\frac{1+\lambda }{2+\lambda }+1}{1+\frac{1+\lambda }{2+\lambda }}=\frac{-1+1/2}{1+1/2}\Rightarrow \lambda =-3$
So the equation of $L_2$ is $y+2x=0\Rightarrow m=-2$

and $812m^2+3=812\times 4+3=3251$