Question

Find the equations of the sides of a triangle having $(4,-1)$ as a vertex, if the lines $x-1=0$ and $x-y-1=0$ are the equations of two internal bisectors of its angles.

Answer 1

Mirror image of point $C(4,-1)$ with respect to $x=1$ lies on $AB$

$x-1=0$------(i)

$\frac{x-4}{1}=\frac{y+1}{0}=\frac{-2(4-1)}{1}$

$\therefore x=-2,y=-1$

$\therefore E(-2,-1)$

Image of $C(4,-1)$ wrt to $x-y-1=0$

$\frac{x-4}{1}=\frac{y+1}{-1}=\frac{-2(4+1-1)}{2}$

$\Rightarrow \frac{x-4}{1}=\frac{y+1}{-1}=-4$

$x=0,y=3\therefore F(0,3)$

$\therefore$ Equation of ${}^{\prime }A{B}^{\prime }$ is $\therefore 2x+y=3$

$\therefore$ put $x=1\therefore y=1,A(1,1)$

Equation of ${}^{\prime }A{C}^{\prime }$ is $2x+3y=5$

$B(\frac{4}{3},\frac{1}{3})$

Equation of ${}^{\prime }B{C}^{\prime }$ is $x+2y=2$

Hence, this is the answer.