+299 Determine the coordinates of the point $P$ on the line $y=-x+6$ such that $P$ is equidistant from the points $A(10,-12)$ and $O(2,8)$ (that is, so that $PA=PO$). Express your answer as an ordered pair $(a,b)$.
+129820 Let P have the coordinates ( x , -x + 6)
PA^2 = ( x - 10)^2 + (- x + 6 - -12)^2 = (x -10)^2 + ( -x + 18)^2
PO^2 = ( x - 2)^2 + ( -x + 6 - 8)^2 = (x -2^2 + ( -x - 2)^2
We want PA^2 = PO^2....so....
(x -10)^2 + ( -x + 18)^2 = ( x -2)^2 + (-x -2)^2
x^2 -20x + 100 + x^2 - 36x + 324 = x^2 -4x + 4 + x^2 +4x + 4
-56x + 424 = 8
-56x = -416
x = 416/56 = 52/7
y = - 52/7 + 6 = -10/7
P = (52/7, -10/7)
