Question

The locus of the point equidistant from the points $(a+b,a-b)$ and $(a-b,a+b)$ is

• A. $bx-ay=0$
• B. $bx+ay=0$
• C. $ax-by=0$
• D. $x-y=0$

Answer 1

Answer: (D)

$x-y=0$

Let the point be $P(x,y)$
Then
$\sqrt{(x-(a+b){)}^2+(y-(a-b){)}^2}=\sqrt{(x-(a-b){)}^2+(y-(a+b){)}^2}$
$(x-(a+b){)}^2+(y-(a-b){)}^2=(x-(a-b){)}^2+(y-(a+b){)}^2$
$(x-(a+b){)}^2-(x-(a-b){)}^2=(y-(a+b){)}^2-(y-(a-b){)}^2$
$(2x-2a)(a-b-(a+b\left)\right)=(2y-2a)(a-b-(a+b\left)\right)$
$2(x-a)(-2b)=2(y-a)(-2b)$
$x-a=y-a$
or
$x-y=0$ is the required equation.