The polar equation of the line perpendicular | vedclass

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Question

(A) The given polar equation of the line is $\sin 	heta - \cos 	heta = \frac{1}{r}$.Multiplying by $r$,we get $r \sin 	heta - r \cos 	heta = 1$.Us

Vedclass · Accepted Answer

$\sin 	heta + \cos 	heta = \frac{\sqrt{3} + 1}{r}$

Vedclass · Answer

(A) The given polar equation of the line is $\sin 	heta - \cos 	heta = \frac{1}{r}$.Multiplying by $r$,we get $r \sin 	heta - r \cos 	heta = 1$.Using the conversion $x = r \cos 	heta$ and $y = r \sin 	heta$,the Cartesian equation is $y - x = 1$,or $x - y + 1 = 0$.The slope of this line is $m_1 = 1$.The line perpendicular to this line will have a slope $m_2 = -1$.The point given is $\left(2, \frac{\pi}{6}ight)$. Converting to Cartesian coordinates:$x = 2 \cos \left(\frac{\pi}{6}ight) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$$y = 2 \sin \left(\frac{\pi}{6}ight) = 2 \cdot \frac{1}{2} = 1$So the point is $(\sqrt{3}, 1)$.The equation of the line with slope $-1$ passing through $(\sqrt{3}, 1)$ is:$y - 1 = -1(x - \sqrt{3})$$y - 1 = -x + \sqrt{3}$$x + y = \sqrt{3} + 1$Converting back to polar form using $x = r \cos 	heta$ and $y = r \sin 	heta$:$r \cos 	heta + r \sin 	heta = \sqrt{3} + 1$$r(\sin 	heta + \cos 	heta) = \sqrt{3} + 1$$\sin 	heta + \cos 	heta = \frac{\sqrt{3} + 1}{r}$

The polar equation of the line perpendicular to the line $\sin \theta - \cos \theta = \frac{1}{r}$ and passing through the point $\left(2, \frac{\pi}{6}\right)$ is

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