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 intersecting 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

The correct option is 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 distance $\frac{1}{2}$ from the origin

$\therefore \left|\frac{k}{\sqrt{2}}\right|=\frac{1}{2}\Rightarrow k=\pm \frac{1}{\sqrt{2}}$

Since $BC$ and the required line are on the same side of the origin

$\therefore k=\pm \frac{1}{\sqrt{2}}$

Hence, the required line is $x+y+\frac{1}{\sqrt{2}}=0$