Question:
If \(x\)-coordinate of a point \(P\) on the line joining the points \(Q(2,2,1)\) and \(R(5,2,-2)\) is \(4\), then the \(y\)-coordinate of \(P\) is
If \(x\)-coordinate of a point \(P\) on the line joining the points \(Q(2,2,1)\) and \(R(5,2,-2)\) is \(4\), then the \(y\)-coordinate of \(P\) is
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For a line joining two points in 3D geometry, first form the parametric equations and then substitute the given coordinate condition to find the unknown coordinates.
Updated On: Jun 22, 2026
- \(-\dfrac{1}{2}\times\) (x-coordinate of \(P\))
- \(-2\times\) (z-coordinate of \(P\))
- \(2\times\) (z-coordinate of \(P\))
- Sum of \(x\) and \(z\) coordinates of \(P\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the parametric form of the line joining \(Q\) and \(R\).
The points are
\[ Q(2,2,1),\qquad R(5,2,-2) \] Direction ratios of the line are
\[ (5-2,\;2-2,\;-2-1) \] \[ =(3,0,-3) \] Therefore, the parametric equations of the line are
\[ x=2+3t \] \[ y=2 \] \[ z=1-3t \]
Step 2: Use the given \(x\)-coordinate.
It is given that the \(x\)-coordinate of point \(P\) is \(4\).
So,
\[ 2+3t=4 \] \[ 3t=2 \] \[ t=\frac{2}{3} \]
Step 3: Find the coordinates of \(P\).
Substitute \(t=\dfrac{2}{3}\):
For \(y\),
\[ y=2 \] For \(z\),
\[ z=1-3\left(\frac{2}{3}\right) \] \[ z=1-2 \] \[ z=-1 \] Thus,
\[ P=(4,2,-1) \]
Step 4: Compare with the options.
We have
\[ y=2 \] and
\[ z=-1 \] Therefore,
\[ -2\times z=-2(-1)=2 \] Hence,
\[ y=-2z \]
Step 5: Final conclusion.
Thus, the \(y\)-coordinate of \(P\) is
\[ \boxed{-2\times \text{(z-coordinate of }P)} \]
The points are
\[ Q(2,2,1),\qquad R(5,2,-2) \] Direction ratios of the line are
\[ (5-2,\;2-2,\;-2-1) \] \[ =(3,0,-3) \] Therefore, the parametric equations of the line are
\[ x=2+3t \] \[ y=2 \] \[ z=1-3t \]
Step 2: Use the given \(x\)-coordinate.
It is given that the \(x\)-coordinate of point \(P\) is \(4\).
So,
\[ 2+3t=4 \] \[ 3t=2 \] \[ t=\frac{2}{3} \]
Step 3: Find the coordinates of \(P\).
Substitute \(t=\dfrac{2}{3}\):
For \(y\),
\[ y=2 \] For \(z\),
\[ z=1-3\left(\frac{2}{3}\right) \] \[ z=1-2 \] \[ z=-1 \] Thus,
\[ P=(4,2,-1) \]
Step 4: Compare with the options.
We have
\[ y=2 \] and
\[ z=-1 \] Therefore,
\[ -2\times z=-2(-1)=2 \] Hence,
\[ y=-2z \]
Step 5: Final conclusion.
Thus, the \(y\)-coordinate of \(P\) is
\[ \boxed{-2\times \text{(z-coordinate of }P)} \]
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