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If the centroid of a triangle with vertices $(4, p,-3),(-1,-1,2)$ and $(3,5,-8)$ is given by mid-point of $(1,4,-2)$ and $(q, 2,-4)$, then $p^2+q^2=$
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Verified Answer
The correct answer is:
$34$
Centroid of triangle with vertices
$(4, p,-3),(-1,-1,2),(3,5,-8)$ is
$$
\left(\frac{4-1+3}{3}, \frac{p-1+5}{3}, \frac{-3+2-8}{3}\right)
$$
which is $\left(2, \frac{4+p}{3},-3\right)$.
Mid-point of $(1,4,-2)(q, 2,-4)$ is $\left(\frac{1+q}{2}, 3,-3\right)$
As both points are,
$$
\begin{array}{rlrl}
& & 2 & =\frac{1+q}{2} \text { and } \frac{4+p}{3}=3 \\
\Rightarrow & \quad p & =5, q=3 \\
\therefore \quad & p^2+q^2 & =25+9=34
\end{array}
$$
$(4, p,-3),(-1,-1,2),(3,5,-8)$ is
$$
\left(\frac{4-1+3}{3}, \frac{p-1+5}{3}, \frac{-3+2-8}{3}\right)
$$
which is $\left(2, \frac{4+p}{3},-3\right)$.
Mid-point of $(1,4,-2)(q, 2,-4)$ is $\left(\frac{1+q}{2}, 3,-3\right)$
As both points are,
$$
\begin{array}{rlrl}
& & 2 & =\frac{1+q}{2} \text { and } \frac{4+p}{3}=3 \\
\Rightarrow & \quad p & =5, q=3 \\
\therefore \quad & p^2+q^2 & =25+9=34
\end{array}
$$
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