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# Geometry

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In the coordinate plane, A = (4,-1), B = (6,2), and C = (-1,2). There exists a point Q and a constant K such that for any point P,

$$PA^2 + PB^2 + PC^2 = 3PQ^2 + k.$$
Find the constant K.

Aug 1, 2020

#1
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I worked out so that Q = (-4,2) and k = 26.

Aug 1, 2020
#2
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I am sorry but this is incorrect I have not yet found another answer but I tryed it and it didn't work. Thank you for trying.

Guest Aug 1, 2020
#3
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Let P have co-ordinates (s, t) then

$$\displaystyle PA^{2}+PB^{2}+PC^{2} \\ = (s-4)^{2}+(t+1)^{2}+(s-6)^{2}+(t-2)^{2}+(s+1)^{2}+(t-2)^{2}, \\ \text{which simplifies to} \\ 3s^{2}-18s+53+3t^{2}-6t+9 \\ =3(s^{2}-6s+9)+3(t^{2}-2t+1)+32 \\ =3\{(s-3)^{2}+(t-1)^{2}\}+32.$$

So Q is the point with co-ordinates (3, 1) and k = 32.

Notice that the co-ordinates of Q are the average of the co-ordinates of A, B and C,

i.e. 3 = (4 + 6 - 1)/3, and 1 = (-1 + 2 + 2)/3.

Is that a coincidence ?

Aug 2, 2020