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# Analytic Geo HELP!!!!!

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Let A = (4,-1), B = (6,2), and C = (-1,2). There exists a point X and a constant k such that for any point P:

PA^2 + PB^2 + PC^2 = 3PX^2 + k.

Find the constant k

Mar 19, 2020
edited by Guest  Mar 19, 2020

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Let

$$A = (4,-1),\ B = (6,2)$$, and $$C = (-1,2)$$.
There exists a point $$X$$ and a constant $$k$$ such that for any point $$P$$:
$$PA^2 + PB^2 + PC^2 = 3PX^2 + k$$.
Find the constant $$k$$

My attempt:

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

The constant k is 32

Mar 20, 2020