Question

Find the equation of the bisector of the angle between the lines $x+2y-11=0$ and $3x-6y-1=0$ which contains the point $\left(1,-3\right)$

Answer 1

$x+2y-11=0$ ………..$\left(1\right)$

$3x-6y-1=0$ …………$\left(2\right)$

Point of intersection of $\left(1\right)$ and $\left(2\right)$

$x+2y-11=0$

$3x-6y-1=0$

$3x+6y-33=0$

$3x-6y-1=0$

_________________

$6x=34$

$x=\frac{17}{3}$

$17-6y-1=0$

$6y=16$

$y=\frac{8}{3}$

Equation passing between $(\frac{17}{3},\frac{8}{3})$ and $(1,-3)$

$y-\frac{8}{3}=\frac{-3-\frac{8}{3}}{1+\frac{17}{3}}(x-\frac{17}{3})$

$3y-8=\frac{+17}{+14}(3x-17)$

$42y-112=+51x-289$

$51x-42y-176=0$.