+1808 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 $A = (7,-11),$ $B = (10,13),$ and $C = (18,-22)$, then find the constant k.
+1892 We can use variables to complete this problem.
Let's let P = (x,y)
Plugging this value of P into what we know, we have
\(PA^2 + PB^2 + PC^2 = \\ (x-7)^2 + (y + 11)^2 + (x-10)^2 + (y-13)^2 + (x-18)^2 + ( y + 22)^2\)
\(3 x^2 - 70 x + 3 y^2 + 40 y + 1247 \)
\(3 [ x^2 - 70/3 x + y^2 + 40/3 y ] + 1247 \)
Now, we must complete the square for x and y. Doing this, we get
\(3[ x^2 - 70/3 x + 4900/36 + y^2 + 40/3y + 1600/36 ] + 1247 - 4900/12 -1600/12 \\ 3 [ (x - 70/6]^2 + (y + 40/6)^2 ] + 2116 / 3 \\ Q = (70/6, -40/6 ) = (35/3 , - 20/3)\)
Thus, we have k = 2116 / 3
So our final answer is 2116/3.
Thanks! :)
