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.

Please explain this using analytic geometry. I really want to understand and know this question so that next time when I encounter a similar one, I'll be able to solve it on my own. Thank you.

Guest Jun 20, 2020

#1

#2**+1 **

Lol I almost forgot to reply. I don't know if you did this on purpose or you just made a mistake, but your answer is completely wrong. I don't know how you got it and I SPECIFICALLY asked to explain. Glad I only used your answer as verification and didn't use it immediately. I found a way to do this after I solved it.

Guest Jun 22, 2020

#3**0 **

For those of you (if anyone) who need help on this, here are some hints:

Hint 1: let P be (x,y) and find an expression for PA^2 + PB^2 + PC^2

WARNING: remember what you got is an EXPRESSION and not an equation!!! So don't start to simplify it like it's an equation! (I almost fell for this part)

Hint 2: Simplify the expression and try to resemble whatever you got into the form 3PQ^2 + k, and remember k is a constant!

I hope this helps. :)

Guest Jun 22, 2020

#4**0 **

Oh. And for those of you who are confused about writing an expression for the first part, remember PA is the distance from P to A. How can you find their distance? If you are still confused, try to draw an example which has easy numbers and coordinates to be working with (ex. (1,1), length of 2, etc...)

Hope this helps! :)

Guest Jun 22, 2020