The equation of a line passing through (p cos | vedclass

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Question

(C) The slope of the line $m = 	an(90^\circ + \alpha) = -\cot \alpha = -\frac{\cos \alpha}{\sin \alpha}$.Using the point-slope form $(y - y_1) = m(x

Vedclass · Accepted Answer

$x \cos \alpha + y \sin \alpha = p$

Vedclass · Answer

(C) The slope of the line $m = 	an(90^\circ + \alpha) = -\cot \alpha = -\frac{\cos \alpha}{\sin \alpha}$.Using the point-slope form $(y - y_1) = m(x - x_1)$:$(y - p \sin \alpha) = -\frac{\cos \alpha}{\sin \alpha}(x - p \cos \alpha)$$y \sin \alpha - p \sin^2 \alpha = -x \cos \alpha + p \cos^2 \alpha$$x \cos \alpha + y \sin \alpha = p(\cos^2 \alpha + \sin^2 \alpha)$Since $\cos^2 \alpha + \sin^2 \alpha = 1$,the equation is $x \cos \alpha + y \sin \alpha = p$.

The equation of a line passing through $(p \cos \alpha, p \sin \alpha)$ and making an angle $(90^\circ + \alpha)$ with the positive direction of the $X$-axis is:

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