The perpendicular distance from the point (1, | vedclass

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(D) Given points $(1, \pi)$,$(1, 0^{\circ})$ and $(1, \frac{\pi}{2})$ are in polar form.Converting these to Cartesian coordinates $(x, y) = (r \cos

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$\sqrt{2}$

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(D) Given points $(1, \pi)$,$(1, 0^{\circ})$ and $(1, \frac{\pi}{2})$ are in polar form.Converting these to Cartesian coordinates $(x, y) = (r \cos 	heta, r \sin 	heta)$:$(1, \pi) ightarrow (1 \cdot \cos \pi, 1 \cdot \sin \pi) = (-1, 0)$$(1, 0^{\circ}) ightarrow (1 \cdot \cos 0^{\circ}, 1 \cdot \sin 0^{\circ}) = (1, 0)$$(1, \frac{\pi}{2}) ightarrow (1 \cdot \cos \frac{\pi}{2}, 1 \cdot \sin \frac{\pi}{2}) = (0, 1)$Now,the equation of the line passing through $(1, 0)$ and $(0, 1)$ is given by the intercept form $\frac{x}{a} + \frac{y}{b} = 1$ or slope-point form:$y - 0 = \frac{1 - 0}{0 - 1}(x - 1)$ $\Rightarrow y = -x + 1$ $\Rightarrow x + y - 1 = 0$.The perpendicular distance $d$ from the point $(-1, 0)$ to the line $x + y - 1 = 0$ is calculated using the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$:$d = \frac{|1(-1) + 1(0) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

The perpendicular distance from the point $(1, \pi)$ to the line joining $(1, 0^{\circ})$ and $(1, \frac{\pi}{2})$ (in polar coordinates) is

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