+1855 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.
+1950 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
PA2+PB2+PC2=(x−7)2+(y+11)2+(x−10)2+(y−13)2+(x−18)2+(y+22)2
3x2−70x+3y2+40y+1247
3[x2−70/3x+y2+40/3y]+1247
Now, we must complete the square for x and y. Doing this, we get
3[x2−70/3x+4900/36+y2+40/3y+1600/36]+1247−4900/12−1600/123[(x−70/6]2+(y+40/6)2]+2116/3Q=(70/6,−40/6)=(35/3,−20/3)
Thus, we have k = 2116 / 3
So our final answer is 2116/3.
Thanks! :)
