the equation of the locus of the points equidistant from the point a(-2,3) and b(6,-5) is ?

a (-2,3)
b (6,-5)
point be (h, k)
(h+2)^ 2 +(k-3)^2 = (h-6)^2 +(k+5)^2
h-k-3=0
therefore,
locus:
x-y-3=0