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

The point B is (2, 1) (B is image of a about lilne y = x). Image of A(1, 2) in the line $x–2y+1=0$ is given by $\frac{x-1}{1}=\frac{y-2}{2}=\frac{4}{5}$

$\therefore$Coordinates of the point are $\left(\frac{9}{5},\frac{2}{5}\right)$ 
Since this point lies on BC.

$\therefore$equation of BC is 3x – y – 5 = 0

$\therefore$a + b = 2