Question

Find the locus of a point which is equidistant from the point (3,4) and (5,-2).

Answer 1

Let the point be $(x,y)$

As it is equidistant from the two points, by distance formula

$\sqrt{(x-2{)}^2+(y-3{)}^2}=\sqrt{(y-5{)}^2+(x-4{)}^2}$

$(x-2{)}^2+(y-3{)}^2=(x-4{)}^2+(y-5{)}^2$

$x^2+4-4x+y^2+9-6y=x^2+16-8x+y^2+25-10y$

$13-4x-6y=41-8x-10y$

$-4x+8x-6y+10y=41-13$

$4x+4y=28$

$x+y=7$