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.

Weiart000 Sep 22, 2020

#1**+2 **

We are given:

A = (4, 1)

B = (6, 2)

C = (-1, 2)

And we can let:

P = (x, y)

By the Pythagorean Theorem/distance formula we can say:

PA^{2} = (x - 4)^{2} + (y - 1)^{2}

PB^{2} = (x - 6)^{2} + (y - 2)^{2}

PC^{2} = (x + 1)^{2} + (y - 2)^{2}

Then....

PA^{2} + PB^{2} + PC^{2} = (x - 4)^{2} + (y - 1)^{2} + (x - 6)^{2} + (y - 2)^{2} + (x + 1)^{2} + (y - 2)^{2}

Expand each term, then combine like terms to get:

PA^{2} + PB^{2} + PC^{2} = 3x^{2} - 18x + 3y^{2} - 10y + 62

(I used this to do that because I am lazy.)

Next we want to get the expression on the right side of the equation into the form:

3PQ^{2} + k

which is:

3[ (x - something)^{2} + (y - something)^{2} ] + k

To do that, we need to complete the squares of the x terms and the y terms.

PA^{2} + PB^{2} + PC^{2} = 3(x^{2} - 6x) + 3(y^{2} - \(\frac{10}{3}\)y) + 62

PA^{2} + PB^{2} + PC^{2} = 3(x^{2} - 6x + 9 - 9) + 3(y^{2} - \(\frac{10}{3}\)y + \(\frac{25}{9}\) - \(\frac{25}{9}\)) + 62

PA^{2} + PB^{2} + PC^{2} = 3( (x - 3)^{2} - 9) + 3( (y - \(\frac53\))^{2} - \(\frac{25}{9}\)) + 62

PA^{2} + PB^{2} + PC^{2} = 3(x - 3)^{2} - 27 + 3(y - \(\frac53\))^{2} - \(\frac{25}{3}\) + 62

PA^{2} + PB^{2} + PC^{2} = 3(x - 3)^{2} + 3(y - \(\frac53\))^{2} - \(\frac{25}{3}\) + 62 - 27

PA^{2} + PB^{2} + PC^{2} = 3[ (x - 3)^{2} + (y - \(\frac53\))^{2} ] - \(\frac{25}{3}\) + 62 - 27

PA^{2} + PB^{2} + PC^{2} = 3[ (x - 3)^{2} + (y - \(\frac53\))^{2} ] + \(\frac{80}{3}\)

(Check)

Now it is in the desired form, and so we can pick out that k = \(\frac{80}{3}\)

BTW, I came across this answer. It works out a bit more cleanly if A = (4, -1). Just wanted to mention this in case that is what you meant.

hectictar Sep 22, 2020