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MHT CET · Maths · Three Dimensional Geometry

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. A \(\left(-2,-2, \frac{3}{2}\right)\)
  2. B \(\left(2,2, \frac{3}{2}\right)\)
  3. C \(\left(\frac{3}{2}, 2,-2\right)\)
  4. D \(\left(\frac{3}{2},-2,-2\right)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\left(\frac{3}{2},-2,-2\right)\)

Step-by-step Solution

Detailed explanation

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