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# If someone could CORRECTLY help me on this please

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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.

Sep 22, 2020

#1
+9041
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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:

PA2  =   (x - 4)2  +  (y - 1)2

PB2  =   (x - 6)2  +  (y - 2)2

PC2  =   (x + 1)2  +  (y - 2)2

Then....

PA2 + PB2 + PC2   =   (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:

PA2 + PB2 + PC2   =   3x2  -  18x  +  3y2  -  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:

3PQ2  +  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.

PA2 + PB2 + PC2   =   3(x2  -  6x)  +  3(y2  -  $$\frac{10}{3}$$y)  +  62

PA2 + PB2 + PC2   =   3(x2  -  6x + 9 - 9)  +  3(y2  -  $$\frac{10}{3}$$y + $$\frac{25}{9}$$  -  $$\frac{25}{9}$$)  +  62

PA2 + PB2 + PC2   =   3( (x - 3)2 - 9)  +  3( (y - $$\frac53$$)2  -  $$\frac{25}{9}$$)  +  62

PA2 + PB2 + PC2   =   3(x - 3)2 - 27  +  3(y - $$\frac53$$)2  -  $$\frac{25}{3}$$  +  62

PA2 + PB2 + PC2   =   3(x - 3)2  +  3(y - $$\frac53$$)2  -  $$\frac{25}{3}$$  +  62 - 27

PA2 + PB2 + PC2   =   3[ (x - 3)2  +  (y - $$\frac53$$)2 ]   -  $$\frac{25}{3}$$  +  62 - 27

PA2 + PB2 + PC2   =   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.

Sep 22, 2020
#2
+232
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Wow thank you so much! I posted this question before and some random guest answered incorrectly, without any work either. So this helped me so much thanks!

Weiart000  Sep 23, 2020