MHT CET · Maths · Three Dimensional Geometry
A point on \(X O Z\) -plane divides the join of \((5,-3,-2)\) and \((1,2,-2)\) at
- A \(\left(\frac{13}{5}, 0,-2\right)\)
- B \(\left(\frac{13}{5}, 0,2\right)\)
- C \((5,0,2)\)
- D \((5,0,-2)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{13}{5}, 0,-2\right)\)
Step-by-step Solution
Detailed explanation
Let point \(P(x, y, z)\) divides the line joining the points \(A\) and \(B\) in the ratio \(m: 1\).

Since, point \(P\) is on \(X O Z\) -plane \(\therefore y\) coordinate \(=0\) \(\Rightarrow \quad \frac{2 m-3}{m+1}=0\)
\(\Rightarrow \quad m=\frac{3}{2}\)
Now, \(x=\frac{3+2 \times 5}{3+2}=\frac{13}{5}\)
and \(z=\frac{3 \times(-2)+2 \times(-2)}{5}=-2\)
\(\therefore\) Required point is \(\left(\frac{13}{5}, 0,-2\right)\).

Since, point \(P\) is on \(X O Z\) -plane \(\therefore y\) coordinate \(=0\) \(\Rightarrow \quad \frac{2 m-3}{m+1}=0\)
\(\Rightarrow \quad m=\frac{3}{2}\)
Now, \(x=\frac{3+2 \times 5}{3+2}=\frac{13}{5}\)
and \(z=\frac{3 \times(-2)+2 \times(-2)}{5}=-2\)
\(\therefore\) Required point is \(\left(\frac{13}{5}, 0,-2\right)\).
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