N(3, -4) is the foot of the perpendicular dra | vedclass

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(C) Let the origin be $O(0, 0)$ and the foot of the perpendicular be $N(3, -4)$.The slope of the line segment $ON$ is $m_{ON} = \frac{-4 - 0}{3 - 0} =

Vedclass · Accepted Answer

$3x - 4y - 25 = 0$

Vedclass · Answer

(C) Let the origin be $O(0, 0)$ and the foot of the perpendicular be $N(3, -4)$.The slope of the line segment $ON$ is $m_{ON} = \frac{-4 - 0}{3 - 0} = -\frac{4}{3}$.Since the line $L$ is perpendicular to $ON$,the slope of line $L$ $(m_L)$ is given by $m_L 	imes m_{ON} = -1$.$m_L 	imes (-\frac{4}{3}) = -1 \Rightarrow m_L = \frac{3}{4}$.The line $L$ passes through the point $N(3, -4)$ and has a slope $m_L = \frac{3}{4}$.Using the point-slope form,the equation of the line $L$ is $y - y_1 = m(x - x_1)$.$y - (-4) = \frac{3}{4}(x - 3)$$4(y + 4) = 3(x - 3)$$4y + 16 = 3x - 9$$3x - 4y - 25 = 0$.

$N(3, -4)$ is the foot of the perpendicular drawn from the origin to a line $L$. Then the equation of the line $L$ is

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