+0

# coordinates

0
101
3

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.

Jan 11, 2021

#1
+303
-1

k = QA^2+QB^2+QC^2 = 32, where Q is the barycentre (3, 1) of ABC.

To see this, replace PA by PG+QA etc., expand the squares and use distributivity and the defining property of the barycenter QA+QB+QC=0.

You can also let P be (x, y) and check that PA^2+PB^2+PC^2 = (x-4)^2+(y+1)^2 + … can be transformed into 3[(x-3)^2 + (y-1)^2] + 32.

Jan 11, 2021
#2
+266
0

Hey... haven't I told you not to plagiarize??? It is bad, and you copied someone else's answer, here is the real answer: https://www.quora.com/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-What-is-the-constant-k

DewdropDancer  Jan 11, 2021
#3
+25658
+1

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.

My attempt:

$$\small{ \begin{array}{|rcll|} \hline \mathbf{PA^2 + PB^2 + PC^2} &=& \mathbf{3PQ^2 + k} \\ \boxed{PA = P-A\\ PB=P-B\\PC=P-C\\ PQ=P-Q } \\ (P-A)^2 + (P-B)^2 + (P-C)^2 &=& 3(P-Q)^2 + k \\ P^2-2PA+A^2 + P^2-2PB+B^2 + P^2-2PC+C^2 &=& 3\left(P^2-2PQ+Q^2 \right)+ k \\ 3P^2+A^2+B^2+C^2-2P(A+B+C) &=& 3P^2-6PQ+3Q^2 + k \\ A^2+B^2+C^2-2P(A+B+C) &=& k+3Q^2-6PQ \\ \boxed{A=\dbinom{4}{-1},\ B=\dbinom{6}{2},\ C=\dbinom{-1}{2} } \\ \dbinom{4}{-1}\dbinom{4}{-1}+\dbinom{6}{2}\dbinom{6}{2}+\dbinom{-1}{2}\dbinom{-1}{2}-2P\Bigg(\dbinom{4}{-1}+\dbinom{6}{2}+\dbinom{-1}{2}\Bigg) &=& k+3Q^2-6PQ \\ (16+1)+(36+4)+(1+4)-2P\Bigg(\dbinom{4+6-1}{-1+2+2}\Bigg) &=& k+3Q^2-6PQ \\ 62-2P \dbinom{9}{3} &=& k+3Q^2-6PQ \\ 62-2P \dbinom{3*3}{3*1} &=& k+3Q^2-6PQ \\ 62-2*3P \dbinom{3}{1} &=& k+3Q^2-6PQ \\ 62-6P \dbinom{3}{1} &=& k+3Q^2-6PQ \\ \text{compare} && \text{compare} \\ {\color{red}62}-6P {\color{blue}\dbinom{3}{1}} &=& {\color{red}k+3Q^2}-6P{\color{blue}Q} \\ \hline \mathbf{{\color{blue}Q}} &=& \mathbf{{\color{blue}\dbinom{3}{1}}} \\\\ {\color{red}k+3Q^2} &=& {\color{red}62} \\ k+3\dbinom{3}{1}\dbinom{3}{1} &=& 62 \\ k+3(3*3+1*1) &=& 62 \\ k+3*10 &=& 62 \\ k+30 &=& 62 \\ k &=& 62-30 \\ \mathbf{k} &=& \mathbf{32} \\ \hline \end{array} }$$

Jan 11, 2021