Question

The vertices of a $△OBC$ are $O(0,0),B(-3,-1)$ and $C(-1,-3).$

The equation of a line parallel to $BC$ and interscting sides $OB$ and $OC$ whose distance from the origin is $\frac{1}{2},$ is?

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

Answer 1

Answer: (A)

$x+y+\frac{1}{\sqrt{2}}=0$

The equation of line $BC$ is $x+y+4=0.$
$\therefore$ Equation of a line parallel to $BC$ is $x+y+k=0.$
This is at a distance $\frac{1}{2}$ from the origin.
$\therefore \left|\frac{k}{\sqrt{2}}\right|=\frac{1}{2}.\therefore k=\pm \frac{1}{\sqrt{2}}$
Since $BC$ and the required line are on the same side of the origin
$\therefore k=\frac{1}{\sqrt{2}}$
Hence, the required line is $x+y+\frac{1}{\sqrt{2}}=0$