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 \[bx=ay\].

Answer:

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.