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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 1 · Set 2 · Q33
Prove that the line through points A (0,-1,-1) and B (4,5,1) intersects the line through points C(3, 9, 4) and D(-4, 4, 4). Hence, write the equation…
Prove that the line through points $\displaystyle \mathrm{A}(0,-1,-1)$ and $\displaystyle \mathrm{B}(4,5,1)$ intersects the line through points C($\displaystyle 3$, $\displaystyle 9$, $\displaystyle 4$) and D(-$\displaystyle 4$, $\displaystyle 4$, $\displaystyle 4$). Hence, write the equation of line passing through the point of intersection of lines AB and CD as well as origin.
Marking-scheme solution
Lines AB and CD respectively are
$\displaystyle \dfrac{x}{2}=\dfrac{y+1}{3}=\dfrac{z+1}{1}$ and
$\displaystyle \dfrac{x-3}{-7}=\dfrac{y-9}{-5}=\dfrac{z-4}{0}$
Any point on line AB is $\displaystyle (2\lambda, 3\lambda-1, \lambda-1)$
Any point on line CD is $\displaystyle (-7\mu+3, -5\mu+9, 4)$
For the lines to intersect, we must have some $\displaystyle \lambda$ and $\displaystyle \mu$, for which these coordinates must coincide, i.e., we must have $\displaystyle \lambda-1 = 4, 2\lambda = -7\mu+3, 3\lambda-1 = -5\mu+9$, for some $\displaystyle \lambda$ and $\displaystyle \mu$
The first two equations, when solved, give us $\displaystyle \lambda = 5$ and $\displaystyle \mu = -1$. These values satisfy the third equation. Hence, the lines intersect.
The point of intersection is $\displaystyle (10, 14, 4)$
The required equation is $\displaystyle \dfrac{x}{10}=\dfrac{y}{14}=\dfrac{z}{4}$ or $\displaystyle \dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}$ or $\displaystyle \dfrac{x-10}{5}=\dfrac{y-14}{7}=\dfrac{z-4}{2}$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.