Question

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

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

Answer 1

Answer: (B)

Equation of line$B C$
$ 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, $D E$ is parallel to $B C]$
$ \Rightarrow \sqrt{2} y=-\sqrt{2} x+1 $
$\Rightarrow \sqrt{2} y+\sqrt{2} x-1=0$
$\Rightarrow \sqrt{2} x+\sqrt{2} y-1=0$
$\Rightarrow 2 x+2 y-\sqrt{2}=0$