Question

Let $A B C$ be a triangle. Let $A$ be the point $(1,2) . y=x$ is the perpendicular bisector of $A B$ and $x-2 y+$ $1=0$ is the angle bisector of $\angle C$. If equation of $B C$ is given by $ax +b y-5=0$, then find the value of $a+b$.

Answer 1

The point $B$ is $(2,1)$
Image of $A (1,2)$ in the line $x -2 y +1=0$ is given
by $\frac{x-1}{1}=\frac{y-2}{-2}=\frac{4}{5}$
$\therefore$ coordinate of the point are $\left(\frac{9}{5}, \frac{2}{5}\right)$
Since this point lies on $BC$.
$\therefore$ equation of $BC$ is $3 x - y -5=0$
$\therefore a+ b=2$