Question

The equation of the line belonging to the family of lines $(x+y)+\lambda (2x-y+1)=0$ and farthest from point $(1,-3)$ is

• A. $6x+15y+7=0$
• B. $6x+15y-7=0$
• C. $6x-15y+7=0$
• D. $6x-15y-7=0$

Answer 1

The correct option is C $6x-15y+7=0$

Given: $Q=(1,-3)$ and the lines are
$x+y=02x-y+1=0\Rightarrow 2x+x+1=0\Rightarrow x=-\frac{1}{3}\Rightarrow y=\frac{1}{3}$
So, the lines are concurrent at
$P=(-\frac{1}{3},\frac{1}{3})$
The member of the family that is farthest from $Q$ will be perpendicular to line $PQ$
$m_{PQ}=\frac{[-3-\left(\frac{1}{3}\right)]}{[1+\left(\frac{1}{3}\right)]}\Rightarrow m_{PQ}=-\frac{\frac{10}{3}}{\frac{4}{3}}=-\frac{5}{2}$

So, the slope of required line
$m=\frac{-1}{m_{PQ}}=\frac{2}{5}$
Therefore, the equation of the required line is
$y-\frac{1}{3}=\frac{2}{5}(x+\frac{1}{3})\Rightarrow 15y-5=6x+2\therefore 6x-15y+7=0$