Let P(x,y) be the point. The squares of the distances from P to the given points must be equal so:
(x+2)^2+y^2=(x-2)^2+(y-10)^2; x^2+4x+4+y^2=x^2-4x+4+y^2-20y+100;
-8x-20y+100=0, -2x-5y+25=0, 2x+5y=25 is the locus (straight line) of the equidistant points.
y=-x+1 is the line, and it intersects the locus when 2x+5(-x+1)=25, substituting for y.
Thus, 2x-5x+5=25, -3x=20, so x=-20/3 and y=20/3+1=23/3. So P is the point (-20/3,23/3), because it lies on the locus and the given line.
In the picture, ABC is an isosceles triangle with B(2,10) and C(-2,0) as the given points. M(0,5) is the midpoint of BC, and AB=AC, so AM is the perpendicular from vertex A onto BC. All points on AM (extended) are equidistant from B and C. The red line y=-x+1 meets AM at A(-20/3,23/3). This is a geometrical representation of the problem.
