JEE Main 2025 — Straight Lines Question with Solution
JEE Main 2025 (3 Apr Shift 2)
Question
Let the equation have equal roots. Then the distance of the point from the line is
Choose an option
Show full solutionCorrect option: A
Correct answer
A
Step-by-step explanation
$\begin{aligned}
& \left(\mathrm{x}^2+2 \mathrm{x}\right)(12-\mathrm{k})=2 \\ & \lambda \mathrm{x}^2+2 \lambda \mathrm{x}-2=0 \quad \mathrm{k} \neq 12 \text { Let } 12-\mathrm{k}=\lambda \\ & \mathrm{D}=0 \\ & 4 \lambda^2+8 \lambda=0 \\ & \lambda=0 \text { or } \lambda=-2 \\ & \Rightarrow 12-\mathrm{k}=-2 \\ & \mathrm{k}=14
\end{aligned}\text { So } P\left(k, \frac{k}{2}\right)=(14,7)\mathrm{d}=\left|\frac{3 \times 14+4 \times 7+5}{5}\right|=15$
option (1)
& \left(\mathrm{x}^2+2 \mathrm{x}\right)(12-\mathrm{k})=2 \\ & \lambda \mathrm{x}^2+2 \lambda \mathrm{x}-2=0 \quad \mathrm{k} \neq 12 \text { Let } 12-\mathrm{k}=\lambda \\ & \mathrm{D}=0 \\ & 4 \lambda^2+8 \lambda=0 \\ & \lambda=0 \text { or } \lambda=-2 \\ & \Rightarrow 12-\mathrm{k}=-2 \\ & \mathrm{k}=14
\end{aligned}\text { So } P\left(k, \frac{k}{2}\right)=(14,7)\mathrm{d}=\left|\frac{3 \times 14+4 \times 7+5}{5}\right|=15$
option (1)
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This is a previous-year question from JEE Main 2025, covering the Straight Lines chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.