A line AB passes through the origin (0, 0) an | vedclass

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(C) The given line $CD$ is $3x + 4y + 6 = 0$. Rewriting in slope-intercept form $y = mx + c$,we get $4y = -3x - 6$,which implies $y = -\frac{3}{4}x -

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$4x - 3y = 0$

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(C) The given line $CD$ is $3x + 4y + 6 = 0$. Rewriting in slope-intercept form $y = mx + c$,we get $4y = -3x - 6$,which implies $y = -\frac{3}{4}x - \frac{6}{4}$. The slope of line $CD$ is $m_1 = -\frac{3}{4}$. Since line $AB$ is perpendicular to line $CD$,the product of their slopes must be $-1$. Let the slope of line $AB$ be $m_2$. Then $m_1 	imes m_2 = -1$,so $(-\frac{3}{4}) 	imes m_2 = -1$,which gives $m_2 = \frac{4}{3}$. Line $AB$ passes through the origin $(0, 0)$. Using the point-slope form $(y - y_1) = m(x - x_1)$,we get $(y - 0) = \frac{4}{3}(x - 0)$. This simplifies to $3y = 4x$,or $4x - 3y = 0$.

$A$ line $AB$ passes through the origin $(0, 0)$ and is perpendicular to the line $CD$ given by $3x + 4y + 6 = 0$. The equation of line $AB$ is:

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