Question 992589
Any pt. on perpendicular bisector of a line segment is equidistant from both the end points.
So, Distance from P to A = Distance from P to B
=> {{{sqrt((x+4)^2+(y-7)^2) = sqrt((x-8)^2+(y+13)^2)}}}
Squaring both sides, & simplifying,
{{{x^2+y^2+8x-14y+65 = x^2+y^2-16x+26y+233}}}
=> {{{24x-40y-168 = 0}}}
=> {{{3x-5y-21 = 0}}} represents all such points P as required.