Question

Let in $\Delta ABC$ the coordinates of $A$ are $\left(0 , 0\right).$ Internal angle bisector of $\angle ABC$ is $x-y+1=0$ and mid-point of $BC$ is $\left(- 1 , 3\right).$ Then, the ordinate of $C$ is

Answer 1

Reflection of $A\left(0 , 0\right)$ about the line $x-y+1=0$ is $\left(- 1 , 1\right)$ which lies on $BC$
The equation of $BC$ is $x=-1$ (line passing through $\left(- 1 , 1\right)\&\left(- 1 , 3\right)$ )
$\Rightarrow B$ is the point of intersection of $x=-1$ and $x-y+1=0$
The point $B$ is $\left(- 1 , 0\right)$
Hence, the coordinates of $C$ are $\left(- 1 , 6\right)$ (since, the mid point of $B\&C$ is $\left(- 1 , 3\right)$ )