Question

Consider a family of straight lines $(x+y)+\lambda (2x-y+1)=0$. Then the equation of the straight line belonging to this family that is farthest from $(1,-3)$ is:

• A. $6x+15y+5=0$
• B. $6x-15y+5=0$
• C. $6x+15y+7=0$
• D. $6x-15y+7=0$

Answer 1

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

$(x+y)+\lambda (2x-y+1)=0\Rightarrow L_1+\lambda L_2=0$
Let point of intersection of $L_1$ and $L_2$ be $P$
$x+y=0....\left(1\right)$
$2x-y+1=0....\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$P\equiv (-\frac{1}{3},\frac{1}{3})$
Now every straight line belonging to this family will pass from point $P$
Let $Q\equiv (1,-3)$
The straight line which is farthest from the point $(1,-3)$ and passing through the point $P$ will be perpendicular to $PQ\Rightarrow L\perp PQ$
$\Rightarrow$ Slope of L $\times$ Slope of $PQ=-1$
Slope of $PQ=\frac{-5}{2}$
$\Rightarrow$ Slope of $L=\frac{2}{5}$
We have point $P(-\frac{1}{3},\frac{1}{3})$ and slope of line $L=\frac{2}{5}$
Equation of line $L$ will be
$6x-15y+7=0$