Question

The vertices of triangle $OBC$ are $O(0,0),B(-2,-5),C(-5,-2)$. The equation of the line parallel to $BC$, intersecting the sides $OB$ and $OC$ and whose perpendicular distance from the origin is $\frac{1}{4}$ is

• A. $2\sqrt{2}x+2\sqrt{2}y-1=0$
• B. $2\sqrt{2}x+2\sqrt{2}y+1=0$
• C. $2\sqrt{2}x+2\sqrt{2}y\pm 1=0$
• D. $2\sqrt{2}x-2\sqrt{2}y+1=0$

Answer 1

The correct option is B $2\sqrt{2}x+2\sqrt{2}y+1=0$

Given information can be seen in the diagram below.

$B\equiv (-2,-5)$ and $C\equiv (-5,-2)$
$\Rightarrow$ Slope of$BC=\frac{-2+5}{-5+2}=-1$
$\Rightarrow$ Required line will have slope $=-1$
$\Rightarrow y=-x+c$ is the equation of required line for a constant $c$.
$\Rightarrow x+y-c=0$
It's distance from the origin is $\frac{1}{4}$
$\Rightarrow \frac{-c}{\sqrt{1+1}}=\pm \frac{1}{4}$
$c=\pm \frac{1}{2\sqrt{2}}$
$2\sqrt{2}x+2\sqrt{2}y\pm 1=0$ is equation of line.
But the lines $OB,OC$ are in third quadrant. This line meets both $OB,OC$ and hence it will also be in third quadrant, so that the intercepts on the axes will be negative.
Required equation of line is $2\sqrt{2}x+2\sqrt{2}y+1=0$