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. $33x+12y-7=0$
• B. $33x+12y+7=0$
• C. $12x+33y-7=0$
• D. $12x+33y+7=0$

Answer 1

The correct option is C $12x+33y-7=0$

Given : $(1+2\lambda )x+(1-\lambda )y+\lambda =0$
$\Rightarrow x+y+\lambda (2x-y+1)=0$
Clearly, it represents a family of lines passing through the intersection of the lines $x+y=0$ and $2x-y+1=0$ i.e., the point $(-\frac{1}{3},\frac{1}{3})$


The required line passes through $(-\frac{1}{3},\frac{1}{3})$ and is perpendicular to the line joining $(1,4)$ and $(-\frac{1}{3},\frac{1}{3})$
So, its equation is
$y-\frac{1}{3}=-\frac{1+\frac{1}{3}}{4-\frac{1}{3}}(x+\frac{1}{3})\Rightarrow \frac{3y-1}{3}=-\frac{4}{11}(x+\frac{1}{3})\therefore 12x+33y-7=0$