Question

The vertices of a triangle $OBC$ are $O(0,0)$, $B(-3,-1)$ and $C(-1,-3)$. If the line joining the point $D$ on $OC$ and $E$ on $OB$ is parallel to $BC$ and the perpendicular distance of $O$ from $DE$ is $\frac{1}{2}$, then the equation of $DE$ is

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

Answer 1

Answer: (B)

Equation of line$BC$
$y+1=\frac{-3+1}{-1+3}(x+3)$
$\Rightarrow y+1=-x-3$
$\Rightarrow y=-x-4$
$y=-x+c...(i)$
$y+x-c=0$
$\Rightarrow \left|\frac{-c}{\sqrt{2}}\right|=\frac{1}{2}$
$\Rightarrow \frac{c}{\sqrt{2}}=\frac{1}{2}$
$\Rightarrow c=\frac{1}{\sqrt{2}}$
Now, $y=-x+\frac{1}{\sqrt{2}}$
[ Since, $DE$ is parallel to $BC]$
$\Rightarrow \sqrt{2}y=-\sqrt{2}x+1$
$\Rightarrow \sqrt{2}y+\sqrt{2}x-1=0$
$\Rightarrow \sqrt{2}x+\sqrt{2}y-1=0$
$\Rightarrow 2x+2y-\sqrt{2}=0$