Question

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 find the value of $a+b$.

Answer 1

Answer: 2.00

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