CBSE 2023 · Region 4 · Set 3 · Q32 · 5 marks
(a)Find the image of the point $\displaystyle (2,-1,5)$ in the line \[\begin{gathered} \frac{\mathrm{x}-11}{10}=\frac{\mathrm{y}+2}{-4}=\frac{\mathrm{z}+8}{-11} \\ \text { OR } \end{gathered} \](b)Vertices B and C of $\displaystyle \triangle \mathrm{ABC}$ lie on the line $\displaystyle \frac{\mathrm{x}+2}{2}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}}{4}$. Find the area of $\displaystyle \Delta \mathrm{ABC}$ given that point A has coordinates $\displaystyle (1,-1,2)$ and the line segment BC has length of $\displaystyle 5$ units.
(a)
Find the image of the point $\displaystyle (2,-1,5)$ in the line \[\begin{gathered} \frac{\mathrm{x}-11}{10}=\frac{\mathrm{y}+2}{-4}=\frac{\mathrm{z}+8}{-11} \\ \text { OR } \end{gathered} \]
(b)
Vertices B and C of $\displaystyle \triangle \mathrm{ABC}$ lie on the line $\displaystyle \frac{\mathrm{x}+2}{2}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}}{4}$. Find the area of $\displaystyle \Delta \mathrm{ABC}$ given that point A has coordinates $\displaystyle (1,-1,2)$ and the line segment BC has length of $\displaystyle 5$ units.
Marking-scheme solution
The given line is $\displaystyle \frac{\mathrm{x}-11}{10}=\frac{\mathrm{y}+2}{-4}=\frac{\mathrm{z}+8}{-11}=\lambda$ (say)
Three Dimensional GeometryEquation of a Line in SpaceApplylong_answerhard
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.