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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
avatar+9131 
+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:

 

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. smiley

 Sep 22, 2020
 #2
avatar+232 
+1

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

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