+1439 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 $(a,b)$.
+129881 Let P = (x , 6-x)
We can equate distances (squared)
(10 - x)^2 + ( -12 - (6-x)^2 = ( 2 -x)^2 + (8 - (6-x) )^2
(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
2x^2 - 56x + 424 = 2x^2 + 8
416 = 56x
x = 416 / 56 = 52 / 7
y = 6 - (52/7) = -10/7
(x, y) = (52/7 , -10/7)
