The polar equation of the line perpendicular | vedclass

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Question

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

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

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(A) Given polar equation of the line is $\sin 	heta - \cos 	heta = \frac{1}{r}$.Multiplying by $r$,we get $r \sin 	heta - r \cos 	heta = 1$.In Cartesian coordinates,$x = r \cos 	heta$ and $y = r \sin 	heta$,so the equation becomes $y - x = 1$ or $x - y + 1 = 0$.The slope of this line is $m_1 = 1$.The line perpendicular to this line will have a slope $m_2 = -\frac{1}{m_1} = -1$.The equation of a line with slope $-1$ is $x + y = k$.This line passes through the point $\left(2, \frac{\pi}{6}ight)$.Converting this point to Cartesian coordinates:$x = 2 \cos \left(\frac{\pi}{6}ight) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$$y = 2 \sin \left(\frac{\pi}{6}ight) = 2 \cdot \frac{1}{2} = 1$Substituting $(\sqrt{3}, 1)$ into $x + y = k$,we get $k = \sqrt{3} + 1$.Thus,the Cartesian equation is $x + y = \sqrt{3} + 1$.Converting back to polar form:$r \cos 	heta + r \sin 	heta = \sqrt{3} + 1$$\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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