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 = (5, -3), then find the constant k.
+27 Let P be (x,y)
PA^2 + PB^2 + PC^2
=(x-4)^2+(y+1)^2+(x-6)^2+(y-2)^2+(x+1)^2+(y-2)^2
=3x^2+3y^2-18x-6y+62
=3(x^2+y^2-6x-2y)+62
=3((x-3)^2+(y-1)^2)+32
This shows that if Q=(3,1), PQ^2=(x-3)^2+(y-1)^2
And so PA^2 + PB^2 + PC^2= 3PQ^2+32
k=32
Hope this helped!!!
