Question

The vertices of a triangle OBC are 0(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 point (0 , 0) is 1/2.

Answer 1

B (-3 , -1) and C is (-1 , -3)
slope of BC = $\frac{-3-(-1)}{-1-(-3)}=-1$
Hence any line parallel to BC will have its slope Hence its equation is
y = - x + c or x + y - c = 0
Its distance from origin is 1 /2
$\frac{c}{(-c)}(\sqrt{1+1)}=\pm \frac{1}{2}$
$\therefore c=\pm \frac{\sqrt{2}}{2}$
Required equation is x + y $\pm \sqrt{2}/2$ = 0
Now the lines OB and OC are in 3rd quadrant . This line meet both OB and OC and hence it will also be in 3rd quadrant , so that the intercepts on the axes will be -ive . Therefore we should choose + sign out of $\pm$
Hence the required line is
x + y + $2\sqrt{2}$ / 2 = 0 or 2x + 2y + \sqrt{2} = 0 $