The length of the perpendicular from the poin | vedclass

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(A) The given equation of the line is $y = x 	an \alpha + c$,which can be rewritten as $x 	an \alpha - y + c = 0$.Multiplying by $\cos \alpha$,we ge

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$c \cos \alpha$

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(A) The given equation of the line is $y = x 	an \alpha + c$,which can be rewritten as $x 	an \alpha - y + c = 0$.Multiplying by $\cos \alpha$,we get $x \sin \alpha - y \cos \alpha + c \cos \alpha = 0$.The length of the perpendicular $p$ from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.Substituting the point $(a \cos \alpha, a \sin \alpha)$ and the line equation $x \sin \alpha - y \cos \alpha + c \cos \alpha = 0$:$p = \frac{|(a \cos \alpha) \sin \alpha - (a \sin \alpha) \cos \alpha + c \cos \alpha|}{\sqrt{\sin^2 \alpha + \cos^2 \alpha}}$$p = \frac{|a \sin \alpha \cos \alpha - a \sin \alpha \cos \alpha + c \cos \alpha|}{\sqrt{1}}$$p = |c \cos \alpha|$.Since $c > 0$,the length is $c \cos \alpha$ (assuming $\cos \alpha > 0$ for the distance to be positive).

The length of the perpendicular from the point $(a \cos \alpha, a \sin \alpha)$ upon the straight line $y = x \tan \alpha + c$,where $c > 0$,is:

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