+0  
 
+1
39
3
avatar

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.

 

Thank you in advance

 Aug 1, 2020
 #1
avatar
0

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

 Aug 1, 2020
 #2
avatar
0

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
avatar
0

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

24 Online Users

avatar
avatar
avatar