Consider the lines x(3 +1)+y(7 +2)=17 +5, being a parameter, all passing through a point P…

Question

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

Accepted Answer

$$\begin{aligned}
& x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5 \
& (x+2 y-5)+\lambda(3 x+7 y-17)=0 \
& L_1+\lambda L_2=0 \
& \Rightarrow \quad P \text { is intersection of } L_1 & L_2 \text { i.e. }(1,2) \
& y-2=m(x-1) \
& m x-y+2-m=0 \
& \text { Distance from origin }=\left|\frac{2-m}{\sqrt{1+m^2}}\right|=\max
\end{aligned}$$

$$\begin{aligned}
& \text { For } m=-\frac{1}{2} \
& \therefore L=y-2=\frac{-1}{2}(x-1) \
& L: x+2 y-5=0 \
& \text { Now, } d=\left|\frac{3+12-5}{\sqrt{5}}\right|=\left|\frac{10}{\sqrt{5}}\right| \
& d^2=\frac{100}{5}=20
\end{aligned}$$

Solutions · 1