Points A, B, and c are given in the coordinate plane. There exists a point Q and a constant k such that for any point P,
PA^2 + PB^2+PC^2=3PQ^2 + k
If A = (4, -1), B = (-3, 1), and C = \((8,6)\), then find the constant k.
+9519 Since the equation PA^2 + PB^2+PC^2=3PQ^2 + k holds true for any point P, we may let P = A, P = B, P = C and get 3 equations.
Let Q = (r, s). Then substituting P = A, B, C respectively gives: \(\begin{cases}53 + 65 = 3((r - 4)^2 + (s + 1)^2) + k\\53 + 146 = 3((r + 3)^2 + (s - 1)^2) + k\\65 + 146 = 3((r - 8)^2 + (s - 6)^2) + k\end{cases}\).
Solving this system is possible, but it is very tedious work, so I will do it with Wolfram Alpha:
By the Wolfram Alpha computation, k = 88, r = 3, s = 2.
In case you're wondering how to solve the system of equations, let (1) be the first equation, (2) be the second equation, (3) be the third equation. Consider (1) - (2), (2) - (3), and (1) - (3).
