The vertices of a triangle OBC are (0,0),(-3, | vedclass

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(A) The slope of line $BC$ passing through $(-3, -1)$ and $(-1, -3)$ is $m = \frac{-3 - (-1)}{-1 - (-3)} = \frac{-2}{2} = -1$. Since the required line

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

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(A) The slope of line $BC$ passing through $(-3, -1)$ and $(-1, -3)$ is $m = \frac{-3 - (-1)}{-1 - (-3)} = \frac{-2}{2} = -1$. Since the required line is parallel to $BC$,its equation is of the form $x + y + \lambda = 0$. The distance of this line from the origin $(0, 0)$ is given by $\frac{|\lambda|}{\sqrt{1^2 + 1^2}} = \frac{1}{2}$. This implies $\frac{|\lambda|}{\sqrt{2}} = \frac{1}{2}$,so $|\lambda| = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Thus,$\lambda = \pm \frac{1}{\sqrt{2}}$. For the line to cut $OB$ and $OC$,it must lie between the origin and the line $BC$. The line $BC$ is $x + y + 4 = 0$. Since the origin $(0,0)$ gives $0+0+4 > 0$,we choose $\lambda$ such that the line lies between them,which gives $\lambda = \frac{1}{\sqrt{2}}$. The equation is $x + y + \frac{1}{\sqrt{2}} = 0$,which simplifies to $2x + 2y + \sqrt{2} = 0$.

The vertices of a triangle $OBC$ are $(0,0)$,$(-3,-1)$,and $(-1,-3)$ respectively. The equation of the line parallel to $BC$ which is at a distance of $\frac{1}{2}$ unit from the origin and cuts $OB$ and $OC$ is:

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