CBSE 2025, Set 1

Question

Find the image of the point $(-1, 5, 2)$ in the line $\dfrac{2x-4}{2} = \dfrac{y}{2} = \dfrac{2-z}{3}$. Also find the length of the line segment joining the given point and its image.

Edvaya · Accepted Answer

The correct answer is Subjective. **Rewrite the line:** $\dfrac{2x-4}{2} = \dfrac{x-2}{1}$ and $\dfrac{2-z}{3} = \dfrac{z-2}{-3}$ Line: $\dfrac{x-2}{1} = \dfrac{y}{2} = \dfrac{z-2}{-3} = t$ **General point on line:** $M = (t+2,\ 2t,\ -3t+2)$ **Foot of perpendicular from $A(-1,5,2)$:** Direction vector $\vec{d} = (1, 2, -3)$ $\overrightarrow{AM} = (t+3,\ 2t-5,\ -3t)$ $\overrightarrow{AM} \cdot \vec{d} = 0$: $$(t+3)(1) + (2t-5)(2) + (-3t)(-3) = 0$$ $$t + 3 + 4t - 10 + 9t = 0 \Rightarrow 14t = 7 \Rightarrow t = \frac{1}{2}$$ **Foot of perpendicular:** $M = \left(\dfrac{5}{2},\ 1,\ \dfrac{1}{2}ight)$ **Image $A'$** — $M$ is the midpoint of $A(-1,5,2)$ and $A'$: $$A' = \left(2\cdot\frac{5}{2}-(-1),\ 2(1)-5,\ 2\cdot\frac{1}{2}-2ight) = (6,\ -3,\ -1)$$ $$\boxed{	ext{Image} = (6,\ -3,\ -1)}$$ **Length $AA'$:** $$|AA'| = \sqrt{(6+1)^2+(-3-5)^2+(-1-2)^2} = \sqrt{49+64+9} = \sqrt{122} 	ext{ units}$$

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