If $A(1,4,2)$ and $C(5,-7,1)$ are two vertices of triangle $\mathrm{ABC}$ and $\mathrm{G}\left(\frac{4}{3},…

If $A(1,4,2)$ and $C(5,-7,1)$ are two vertices of triangle $\mathrm{ABC}$ and $\mathrm{G}\left(\frac{4}{3}, 0, \frac{-2}{3}\right)$ is centroid of the triangle $\mathrm{ABC}$, then the mid point of side $\mathrm{BC}$ is
  1. $\left(-2,-2, \frac{3}{2}\right)$
  2. $\left(2,2, \frac{3}{2}\right)$
  3. $\left(\frac{3}{2}, 2,-2\right)$
  4. $\left(\frac{3}{2},-2,-2\right)$

Solution

Let $\mathrm{B} \equiv\left(x_1, y_1, \mathrm{z}_1\right)$ Co-ordinates of centroid $\begin{gathered} \equiv\left(\frac{1+x_1+5}{3}, \frac{4+y_1-7}{3}, \frac{2+\mathrm{z}_1+1}{3}\right) \\ \Rightarrow\left(\frac{4}{3}, 0, \frac{-2}{3}\right) \equiv\left(\frac{6+x_1}{3}, \frac{y_1-3}{3}, \frac{3+\mathrm{z}_1}{3}\right) \\ \Rightarrow x_1=-2, y_1=3, \mathrm{z}_1=-5 \\ \therefore \quad \mathrm{B} \equiv(-2,3,-5) \\ \text { Midpoint side } \mathrm{BC}=\left(\frac{-2+5}{2}, \frac{3-7}{2}, \frac{-5+1}{2}\right) \\ =\left(\frac{3}{2},-2,-2\right) \end{gathered}$

Asked in: MHT CET 2023 (13 May Shift 1)

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