Question

If point $p(x,y)$ is equadist distance from $A(a-b,a+b)B(a+b,b-a)$.Then prove that bx=ay

Answer 1

$p=(x,y)A=(a-b,a+b)B=(a+b,b-a)$

Given $AP=BP$

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

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

$a^2+b^2+x^2-2ab-2ax+2bx+a^2+b^2+y^2+2ab-2ay-2by$

$=a^2+b^2+x^2+2ab-2bx+b^2+a^2+y^2-2ab-2by+2ay$

$2bx-2ay=-2bx+2ay$

$4bbx=4ay$

$\therefore bx=ay$