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.
B (-3 , -1) and C is (-1 , -3)
slope of BC = $$ \dfrac {- \, 3 \, - \, (-1) }{- \, 1\, - \, (- \, 3) } \, = \, - 1$$
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
$$ \dfrac c{(- c )}{( \sqrt{1\, + \, 1)}}\, = \, \pm \, \dfrac {1}{2} $$
$$ \therefore \, \, c \, = \, \pm \, \dfrac {\sqrt{2}}{2}$$
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 $$
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