Q46 / 75 · JEE Main 2025 (3 Apr Shift 2) · Mathematics
Consider the lines $\mathrm{x}(3 \lambda+1)+\mathrm{y}(7 \lambda+2)=17 \lambda+5$, $\lambda$ being a parameter, all passing through a point P . One of these lines (say L) is farthest from the origin. If the distance of $L$ from the point $(3,6)$ is $d$, then the value of $d^2$ is
- Correct answerA$20$
- B$30$
- C$10$
- D$15$
Solution
$\begin{aligned}<br/>& \mathrm{x}(3 \lambda+1)+\mathrm{y}(7 \lambda+2)=17 \lambda+5 \\
& (\mathrm{x}+2 \mathrm{y}-5)+\lambda(3 \mathrm{x}+7 \mathrm{y}-17)=0<br/>\end{aligned}$
intersection of family of lines
$\mathrm{P}(1,2)$
Let $\mathrm{Q}(3,6)$
$\begin{aligned}<br/>& \mathrm{d}=\mathrm{PQ}=\sqrt{2^2+4^2}=\sqrt{20} \\ & \mathrm{~d}^2=20<br/>\end{aligned}$
option (1)
intersection of family of lines
$\mathrm{P}(1,2)$
Let $\mathrm{Q}(3,6)$
$\begin{aligned}<br/>& \mathrm{d}=\mathrm{PQ}=\sqrt{2^2+4^2}=\sqrt{20} \\ & \mathrm{~d}^2=20<br/>\end{aligned}$
option (1)
Hence, the correct answer is option: 'A'.