Mathematics · Straight lines

JEE Main 2025 — 3 April, Evening Shift — Question 36

Consider the lines $x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5$ , $\lambda$ 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

Answer: B

$x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5$

$(x+2 y-5)+\lambda(3 x+7 y-17)=0$ $L_{1}+\lambda L_{2}=0$

$\Rightarrow P$ is intersection of $L_{1} \& L_{2}$ i.e. $(1,2)$ $y-2=m(x-1)$

$m x-y+2-m=0$

Distance from origin $=\left|\frac{2-m}{\sqrt{1+m^{2}}}\right|=\max$

$2 y-4=-x+1$

For $m=-\frac{1}{2}$

$\therefore L=y-2=\frac{-1}{2}(x-1)$

$L: x+2 y-5=0$

Now, $d=\left|\frac{3+12-5}{\sqrt{5}}\right|=\left|\frac{10}{\sqrt{5}}\right|$

$d^{2}=\frac{100}{5}=20$

Answer key and solution verified before publishing.

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