The line passing through left(-1, frac{pi}{2} | vedclass

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Question

(A) Given equation of the line is $\sqrt{3} \sin 	heta + 2 \cos 	heta = \frac{4}{r}$,which can be written as $\sqrt{3} r \sin 	heta + 2 r \cos 	he

Vedclass · Accepted Answer

$2 = \sqrt{3} r \cos 	heta - 2 r \sin 	heta$

Vedclass · Answer

(A) Given equation of the line is $\sqrt{3} \sin 	heta + 2 \cos 	heta = \frac{4}{r}$,which can be written as $\sqrt{3} r \sin 	heta + 2 r \cos 	heta = 4$. Converting to Cartesian coordinates using $x = r \cos 	heta$ and $y = r \sin 	heta$,we get $\sqrt{3} y + 2 x = 4$,or $2x + \sqrt{3} y - 4 = 0$. The slope of this line is $m_1 = -\frac{2}{\sqrt{3}}$. The slope of a line perpendicular to this is $m_2 = -\frac{1}{m_1} = \frac{\sqrt{3}}{2}$. The point $\left(-1, \frac{\pi}{2}ight)$ in polar coordinates corresponds to Cartesian coordinates $(x_1, y_1) = (-1 \cos \frac{\pi}{2}, -1 \sin \frac{\pi}{2}) = (0, -1)$. The equation of the line passing through $(0, -1)$ with slope $\frac{\sqrt{3}}{2}$ is $y - (-1) = \frac{\sqrt{3}}{2}(x - 0)$. $y + 1 = \frac{\sqrt{3}}{2} x$ $\Rightarrow 2y + 2 = \sqrt{3} x$ $\Rightarrow \sqrt{3} x - 2y = 2$. Substituting $x = r \cos 	heta$ and $y = r \sin 	heta$,we get $\sqrt{3} r \cos 	heta - 2 r \sin 	heta = 2$.

The line passing through $\left(-1, \frac{\pi}{2}\right)$ and perpendicular to $\sqrt{3} \sin \theta + 2 \cos \theta = \frac{4}{r}$ is:

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