Question

In a triangle ABC, the bisectors of angles B and C lie long the lines $x=y$ and $y=0$. If A is $(1,2)$, then the equation of line BC is

Answer 1

Reflections of A in the two angle bisectors will lie on the line BC.

Let the images be $A_1$ and $A_2$ respectively.

Formula for image of a point (h, k) in line $ax+by+c=0$ is
$\frac{x-h}{a}=\frac{y-k}{b}=\frac{-2(ah+bk+c)}{a^2+b^2}$
where $(x,y)$ are co-ordinates of image

Image of $A(1,2)\text{in}x-y=0$ is
$\frac{x-1}{1}=\frac{y-2}{-1}=-\frac{2(1+2(-1\left)\right)}{1^2+(-1{)}^2}$
$\Rightarrow x-1=-y+2=1$
$A_1=(2,1)$
$\text{Similarly image of}A(1,2)\text{in}y=0\text{is}(1,-2)$
$A_2=(1,-2)$
$\text{Equation of BC}:y+2=\frac{-2-1}{1-2}(x-1)$
$\Rightarrow 3x-y=5$