Question

If the point $R(x,y)$ is equidistant from the point $P(a+b,a-b)$ and $Q(b-a,a+b)$, then prove that $xa=yb$.

Answer 1

$Distanceofline$ $RP=DistanceoflineRQ{[x-(a+b\left)\right]}^2+[y-(a-b){]}^2=[x-(b-a){]}^2+[y-(a+b){]}^2{x}^2+(a+b{)}^2-2(a+b)x+y^2(a-b{)}^2-2y(a-b)=x^2+(b-a{)}^2-2x(b-a)+y^2+(a+b{)}^2-2y(a+b)-2(a+b)x-2y(a-b)=-2x(b-a)-2y(a+b\left)\right(a+b)a+(a-b)y=(b-a)x+(a+b)yax+bx+ay-by=bx-ax+ay+by2ax=2by\therefore ax=by$