Question

If the point (x, y) is equidistant from the points (a + b, b − a) and (a − b, a + b), prove that bx = ay.

Answer 1

As per the question, we have
$$\begin{gathered} \sqrt{{\left(x-a-b\right)}^2+{\left(y-b+a\right)}^2}=\sqrt{{\left(x-a+b\right)}^2+{\left(y-a-b\right)}^2} \\ \Rightarrow {\left(x-a-b\right)}^2+{\left(y-b+a\right)}^2={\left(x-a+b\right)}^2+{\left(y-a-b\right)}^2\left(\mathrm{Squaring}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow x^2+{\left(a+b\right)}^2-2x\left(a+b\right)+y^2+{\left(a-b\right)}^2-2y\left(a-b\right)=x^2+{\left(a-b\right)}^2-2x\left(a-b\right)+y^2+{\left(a+b\right)}^2-2y\left(a+b\right) \\ \Rightarrow -x\left(a+b\right)-y\left(a-b\right)=-x\left(a-b\right)-y\left(a+b\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow -xa-xb-ay+by=-xa+bx-ya-by \\ \Rightarrow by=bx \end{gathered}$$
Hence, bx = ay.