Question

The vertices of a triangle $OBC$ are $O(0,0),B(-3,-1),C(-1,-3)$, then the equation of the line parallel to the side $BC$ and cutting the sides $OB$ and $OC$ and at a distance $\frac{1}{2}$ from the origin 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: (B)

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

Equation of line parallel to BC is
$y=-x+c$
Its distance from origin is given as $\frac{1}{2}$
$\Rightarrow \left|\frac{c}{\sqrt{2}}\right|=\frac{1}{2}$
$\Rightarrow c=\pm \frac{1}{\sqrt{2}}$
Since, the line cuts OB and OC, then $c$ must be negative.
Hence, $c=-\frac{1}{\sqrt{2}}$
Hence, the equation becomes $x+y+\frac{1}{\sqrt{2}}=0$