Question

The vertices of $\Delta OBC$ are $O(0,0),B(-3,-1),C(-1,-3)$. Find the equation of the line parallel to $BC$ and intersecting the sides $OB$ and $OC$ and whose perpendicular distance from the origin is $\frac{1}{2}$.

Answer 1

$D(0,0)$ $B(-3,-1)$ $C(-1,-3)$

Equation of the line BC is given by

$y-(-1)=\frac{-3-(-1)}{-1-(-3)}x+3$

$y+1=\frac{-2}{2}(x+3)$

$x+y+4=0$ ………..$\left(1\right)$

line $||$ to BC is given by

$x+y+\lambda =0$ ……..$\left(2\right)$

Length of the $\perp$ from $(0,0)$ on $\left(2\right)$ is

$\frac{\lambda }{\sqrt{2}}=\pm \frac{1}{2}$

$\lambda =\pm \frac{\sqrt{2}}{2}=\pm \frac{1}{\sqrt{2}}$

Choosing $+$ sign as $-$ sign would mean line on the other side of O, i.e., not interesting OB and OC.

Hence the equation of BC is given by

$x+y+\frac{\sqrt{2}}{2}=0$ i.e., $2x+2y+\sqrt{2}=0$.