JEE MAIN - Mathematics (2025 - 3rd April Evening Shift - No. 7)
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
10
20
15
30
Explanation
$$\begin{aligned} & x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5 \\ & (x+2 y-5)+\lambda(3 x+7 y-17)=0 \\ & L_1+\lambda L_2=0 \\ & \Rightarrow \quad P \text { is intersection of } L_1 \& L_2 \text { i.e. }(1,2) \\ & y-2=m(x-1) \\ & m x-y+2-m=0 \\ & \text { Distance from origin }=\left|\frac{2-m}{\sqrt{1+m^2}}\right|=\max \end{aligned}$$
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$$\begin{aligned} & \text { For } m=-\frac{1}{2} \\ & \therefore L=y-2=\frac{-1}{2}(x-1) \\ & L: x+2 y-5=0 \\ & \text { Now, } d=\left|\frac{3+12-5}{\sqrt{5}}\right|=\left|\frac{10}{\sqrt{5}}\right| \\ & d^2=\frac{100}{5}=20 \end{aligned}$$
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