Question

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

Answer 1

The distances of the point $(2,1)$ and $(1,-2)$ from $(x,y)$ are $\sqrt{(x-2{)}^2+(y-1{)}^2}$ and $\sqrt{(x-1{)}^2+(y+2{)}^2}$ units respectively.

According to the problem,

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

or, $(x-2{)}^2+(y-1{)}^2=(x-1{)}^2+(y+2{)}^2$

or, $-4x+4-2y+1=-2x+1+4y+4$

or, $2x+6y=0$

or, $x+3y=0$.