Question

If the points (2, 1) and (1, -2) are equidistant from the point (x, y), show that x + 3y =0.

Answer 1

Let $D_1$ be the distance between $(x,y)$ and $(2,1)$
$D_2$ be the distance between $(x,y)$ and $(1,-2)$
$D_1=\sqrt{(x-2{)}^2+(y-1{)}^2)}$
$D_1=\sqrt{(x^2-4x+4+y^2-2y+1)}$
$D_1=\sqrt{(x^2+y^2-4x-2y+5)}$

$D_2=\sqrt{(x-1{)}^2+(y+2{)}^2)}$
$D_2=\sqrt{(x^2-2x+1+y^2+4y+4)}$
$D_2=\sqrt{(x^2+y^2-2x+4y+5)}$

Here,
$D_1=D_2$
$\sqrt{(x^2+y^2-4x-2y+5)}=\sqrt{(x^2+y^2-2x+4y+5)}$
$-4x-2y+5=-2x+4y+5$
$2x+6y=0$
$x+3y=0$
Hence proved.