The perpendicular from the origin to a line m | vedclass

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Question

(A) Let the origin be $O(0, 0)$ and the given point be $P(-2, 9)$.The slope of the line segment $OP$ is $m_{1} = \frac{9 - 0}{-2 - 0} = -\frac{9}{2}$.

Vedclass · Accepted Answer

$2x - 9y + 85 = 0$

Vedclass · Answer

(A) Let the origin be $O(0, 0)$ and the given point be $P(-2, 9)$.The slope of the line segment $OP$ is $m_{1} = \frac{9 - 0}{-2 - 0} = -\frac{9}{2}$.Since the line is perpendicular to $OP$,its slope $m_{2}$ is given by $m_{2} = -\frac{1}{m_{1}} = -\frac{1}{(-9/2)} = \frac{2}{9}$.The equation of the line passing through $P(-2, 9)$ with slope $m_{2} = \frac{2}{9}$ is given by the point-slope form:$y - y_{1} = m(x - x_{1})$$y - 9 = \frac{2}{9}(x - (-2))$$9(y - 9) = 2(x + 2)$$9y - 81 = 2x + 4$$2x - 9y + 85 = 0$.

The perpendicular from the origin to a line meets it at the point $(-2, 9)$. Find the equation of the line.

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