Question

Find the locus of the point which is equidistant to the points $(2,3)$ and $(4,-1)$.

Answer 1

Let the point be $(x,y)$.

Now the distance of the point from the pints $(2,3)$ and $(4,-1)$ are $\sqrt{(x-2{)}^2+(y-3{)}^2}$ and $\sqrt{(x-4{)}^2+(y+1{)}^2}$.

According to the problem,

$\sqrt{(x-2{)}^2+(y-3{)}^2}=$ $\sqrt{(x-4{)}^2+(y+1{)}^2}$

or, $(x-2{)}^2+(y-3{)}^2=$ $(x-4{)}^2+(y+1{)}^2$

or, $-4x+4-6x+9=-8x+16+2y+1$

or, $4x-8y=4$

or, $x-2y=1$.

This is the locus of the point $(x,y)$.