• # question_answer If the point $P(x,y)$ is equidistant from the points $A(a+b,b-a)$ and $B(a-b,a+b)$. Prove that $bx=ay$.

 Since, P us equidistant from points A and B, $\therefore$                  $PA=PB$ Or,                    ${{(PA)}^{2}}={{(PB)}^{2}}$ ${{(a+b-x)}^{2}}+{{(b-a-y)}^{2}}={{(a-b-x)}^{2}}+{{(a+b-y)}^{2}}$ ${{(a+b)}^{2}}+{{x}^{2}}-2ax-2bx+{{(b-a)}^{2}}+{{y}^{2}}-2by+2ay$$={{(a-b)}^{2}}+{{x}^{2}}-2ax+2bx+{{(a+b)}^{2}}+{{y}^{2}}-2ay-2by$ $-2bx+2ay=2bx-2ay$ $4ay=4bx$ $ay=bx$ $bx=ay$                                  Hence Proved.