The vertices of Delta OBC are (0, 0),(-3, 1), | vedclass

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(D) The slope of $BC$ is $m = \frac{-3 - 1}{-1 - (-3)} = \frac{-4}{2} = -2$. Since the line is parallel to $BC$,its slope is also $-2$. The equation o

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None of these

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(D) The slope of $BC$ is $m = \frac{-3 - 1}{-1 - (-3)} = \frac{-4}{2} = -2$. Since the line is parallel to $BC$,its slope is also $-2$. The equation of the line is $y = -2x + c$,or $2x + y - c = 0$. The distance of this line from the origin $(0, 0)$ is given as $d = 1/2$. Using the perpendicular distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$,we have $\frac{|-c|}{\sqrt{2^2 + 1^2}} = \frac{1}{2}$. $\frac{|c|}{\sqrt{5}} = \frac{1}{2} \implies |c| = \frac{\sqrt{5}}{2} \implies c = \pm \frac{\sqrt{5}}{2}$. Thus,the equation is $2x + y \pm \frac{\sqrt{5}}{2} = 0$,which simplifies to $4x + 2y \pm \sqrt{5} = 0$. Comparing this with the given options,none of them match the derived equation.

The vertices of $\Delta OBC$ are $(0, 0)$,$(-3, 1)$,and $(-1, -3)$ respectively. $A$ line parallel to $BC$ intersects $OB$ and $OC$ at a distance of $1/2$ from $O$. Find the equation of this line.

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