Question

If $A(9,-9),B(1,-3)$ are the ends of the side $AB$ a right angled isosceles triangle, then the third vertex is $(8,-2)$ and $(2,-10)$.

Answer 1

Let the third vertex be $P$

$PA=PB\Rightarrow 4h-3k-38=0$

$PA\perp PB\Rightarrow \frac{k+9}{h-9}.\frac{k+3}{h-1}=-1$

$h^2+k^2-10h+12k+36=0$

Eliminate $k$ between $\left(1\right)$ and $\left(2\right)$ you will have a quadratic in $h$ given $h=2,8\therefore k=-10,-2$.

$\therefore$ Points are $(2,-10)$ and $(8,-2)$.