Question

A family of lines is given by $(1+2\lambda )x+(1-\lambda )y+\lambda =0,\lambda$ being the parameter. The line belonging to this family at the maximum distance from the point $(1,4)$ is

• A. $4x-y+1=0$
• B. $33x+12y+7=0$
• C. $12x+33y=7$
• D. None of these

Answer 1

Answer: (C)

$12x+33y=7$

Here, $x+y+\lambda (2x-y+1)=0$
Since, each line passes through the point of intersection of the lines $x+y=0$ and $2x-y+1=0$, which is $(\frac{-1}{3},\frac{1}{3})$.
Thus slope of the required line, $m=\frac{-1}{\frac{4-\frac{1}{3}}{1+\frac{1}{3}}}=\frac{-4}{11}$
Hence, the required line is $y-\frac{1}{3}=-\frac{4}{11}(x+\frac{1}{3})$

$\Rightarrow \frac{3y-1}{3}=-\frac{4}{11}\left(\frac{3x+1}{3}\right)$

$\Rightarrow 11(3y-1)=-4(3x+1)$

$\Rightarrow 33y-11=-12x-4$

$\Rightarrow 12x+33y=7$