Question

If the point P (x, y) is equidistant from the points $A(a+b,b-a)$ and $B(a-b,a+b)$ then

• A. $ax=by$
• B. $bx=ay$
• C. $ax=-by$
• D. $bx=-ay$

Answer 1

Answer: (A)

$ax=by$

Let $P(x,y)$ be equidistant from the points $A(a+b,b-a)$ and $B(a-b,a+b)$

$\Rightarrow AP=PB$

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

Squaring both sides, we get

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

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

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

$\Rightarrow -2x(a+b)+2y(a-b)=-2x(a-b)-2y(a+b)$

$\Rightarrow x(a+b)-y(a-b)=x(a-b)+y(a+b)$

$\Rightarrow ax+bx-ay+by=ax-bx+ay+by$

$\Rightarrow 2bx=2ay$

or $ax=by$