Question

If $P(x,y)$ is equidistant from $A(a+b,b-a)$ and $B(a-b,a+b)$, show that $bx=ay$.

Answer 1

$PA=PB\Rightarrow {PA}^2={PB}^2$

$\Rightarrow {(a+b-x)}^2+{(b-a-y)}^2={(a-b-x)}^2+{(a+b-y)}^2$

$\Rightarrow {(a+b-x)}^2-{(a-b-x)}^2={(a+b-y)}^2-{(b-a-y)}^2$

$2(a-x).2b=2(b-y)2a$

$\Rightarrow ab-bx=ab-ay$

$\Rightarrow bx=ay$ Proved.