O(0,0), A(-3,-1) and B(-1,-3) are the vertice | vedclass

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(B) The vertices are $O(0,0), A(-3,-1)$,and $B(-1,-3)$.Slope of $OB = \frac{-3-0}{-1-0} = 3$.Equation of line $OB$ is $y = 3x$,or $3x - y = 0$.Since $

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$21x-7y+32=0$

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(B) The vertices are $O(0,0), A(-3,-1)$,and $B(-1,-3)$.Slope of $OB = \frac{-3-0}{-1-0} = 3$.Equation of line $OB$ is $y = 3x$,or $3x - y = 0$.Since $AD \perp OB$,the slope of $AD$ is $-\frac{1}{3}$.Equation of line $AD$ passing through $A(-3,-1)$ is $y - (-1) = -\frac{1}{3}(x - (-3))$,which simplifies to $y + 1 = -\frac{1}{3}(x + 3)$,or $x + 3y + 6 = 0$.To find $D$,the intersection of $OB$ $(y=3x)$ and $AD$ $(x+3y+6=0)$:$x + 3(3x) + 6 = 0$ $\Rightarrow 10x = -6$ $\Rightarrow x = -\frac{3}{5}$.$y = 3(-\frac{3}{5}) = -\frac{9}{5}$. So,$D = (-\frac{3}{5}, -\frac{9}{5})$.Point $P$ divides $AD$ in the ratio $3:4$. Using the section formula:$P = \left( \frac{3(-\frac{3}{5}) + 4(-3)}{3+4}, \frac{3(-\frac{9}{5}) + 4(-1)}{3+4} \right) = \left( \frac{-\frac{9}{5} - 12}{7}, \frac{-\frac{27}{5} - 4}{7} \right) = \left( \frac{-69}{35}, \frac{-47}{35} \right)$.Line $L$ is parallel to $OB$ $(y=3x)$,so its equation is $y = 3x + k$.Since $L$ passes through $P(-\frac{69}{35}, -\frac{47}{35})$:$-\frac{47}{35} = 3(-\frac{69}{35}) + k \Rightarrow k = \frac{-47 + 207}{35} = \frac{160}{35} = \frac{32}{7}$.Equation $L: y = 3x + \frac{32}{7}$ $\Rightarrow 7y = 21x + 32$ $\Rightarrow 21x - 7y + 32 = 0$.

$O(0,0), A(-3,-1)$ and $B(-1,-3)$ are the vertices of a $\triangle OAB$. $P$ is a point on the perpendicular $AD$ drawn from $A$ on $OB$ such that $\frac{AP}{PD}=\frac{3}{4}$. Then the equation of the line $L$ parallel to $OB$ and passing through $P$ is:

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