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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.

 May 12, 2022
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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!!!

 May 12, 2022

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