Question

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

Answer 1

Given that, Distance of point (x,y) from (2,1)

$D_1=\sqrt{{(x-2)}^2+{(y-1)}^2}$

Distance of point (x,y) from (1,-2)

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

Given that,

$D_1=D_2$

$D_1^2=D_2^2$

$x^2+4-4x+y^2+1-2y=x^2+1-2x+y^2+4+4y$

$x+3y=0$

Hence proved