+1234 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 $(a,b)$.
+9673 Let \(P = (x, -x+6)\). We have
Note that \(m_{AO} = \dfrac{8 - (-12)}{2 - 10} = -\dfrac 52\). So the slope of perpendicular bisector of AO is \(\dfrac25\).
Midpoint of AO is \(\left(\dfrac{10+2}2, \dfrac{-12+8}2\right) = (6,-2)\). Since PA = PO, P lies on the perpendicular bisector of AO. Then,
\(\dfrac{(-x+6) - (-2)}{x - 6} = \dfrac{2}5\\ 5(8 - x) = 2(x - 6)\\ 40 - 5x = 2x - 12\\ 52 = 7x\\ x = \dfrac{52}7\)
The point is \(\left(\dfrac{52}7, -\dfrac{52}7 + 6\right) = \left(\dfrac{52}7, -\dfrac{10}7\right)\).
