O(0,0), B(-3,-1), C(-1,-3) are vertices of a | vedclass

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(D) The vertices are $O(0,0), B(-3,-1), C(-1,-3)$.The equation of line $BC$ passing through $(-3,-1)$ and $(-1,-3)$ is $y - (-1) = \frac{-3 - (-1)}{-1

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$1 : 4\sqrt{2} - 1$

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(D) The vertices are $O(0,0), B(-3,-1), C(-1,-3)$.The equation of line $BC$ passing through $(-3,-1)$ and $(-1,-3)$ is $y - (-1) = \frac{-3 - (-1)}{-1 - (-3)}(x - (-3))$,which simplifies to $y + 1 = -1(x + 3)$,or $x + y + 4 = 0$.The altitude from $O(0,0)$ to $BC$ is the perpendicular distance from $O$ to $x + y + 4 = 0$,which is $h = \frac{|0 + 0 + 4|}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.The line $DE$ is $2x + 2y + \sqrt{2} = 0$,which can be written as $x + y + \frac{\sqrt{2}}{2} = 0$,or $x + y + \frac{1}{\sqrt{2}} = 0$.The distance of $DE$ from $O(0,0)$ is $d = \frac{|0 + 0 + \frac{1}{\sqrt{2}}|}{\sqrt{1^2 + 1^2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2}$.The line $DE$ divides the altitude $h$ into two segments of lengths $d$ and $h - d$.The ratio is $d : (h - d) = \frac{1}{2} : (2\sqrt{2} - \frac{1}{2}) = 1 : (4\sqrt{2} - 1)$.

$O(0,0), B(-3,-1), C(-1,-3)$ are vertices of a triangle $OBC$. $D$ is a point on $OC$ and $E$ is a point on $OB$. If the equation of $DE$ is $2x + 2y + \sqrt{2} = 0$,then the ratio in which the line $DE$ divides the altitude of the triangle $OBC$ from $O$ to $BC$ is:

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