+1271 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 $(a,b)$.
+9676 Let P=(x,−x+6). We have
Note that mAO=8−(−12)2−10=−52. So the slope of perpendicular bisector of AO is 25.
Midpoint of AO is (10+22,−12+82)=(6,−2). Since PA = PO, P lies on the perpendicular bisector of AO. Then,
(−x+6)−(−2)x−6=255(8−x)=2(x−6)40−5x=2x−1252=7xx=527
The point is (527,−527+6)=(527,−107).
