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

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Question

(A) The given equation is $\sqrt{3} \sin 	heta + 2 \cos 	heta = \frac{4}{r}$.Multiplying by $r$,we get $\sqrt{3} r \sin 	heta + 2 r \cos 	heta = 4

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

Vedclass · Answer

(A) The given equation is $\sqrt{3} \sin 	heta + 2 \cos 	heta = \frac{4}{r}$.Multiplying by $r$,we get $\sqrt{3} r \sin 	heta + 2 r \cos 	heta = 4$.Using polar coordinates $x = r \cos 	heta$ and $y = r \sin 	heta$,the equation becomes $\sqrt{3} y + 2 x = 4$,or $2x + \sqrt{3}y - 4 = 0$.The slope of this line is $m_1 = -\frac{2}{\sqrt{3}}$.$A$ line perpendicular to this will have a slope $m_2 = \frac{\sqrt{3}}{2}$.The equation of a line with slope $m_2$ passing through the point $(x_1, y_1) = (-1, \frac{\pi}{2})$ in polar form is tricky,so we convert the point to Cartesian coordinates: $x = r \cos 	heta = -1 \cos(\frac{\pi}{2}) = 0$ and $y = r \sin 	heta = -1 \sin(\frac{\pi}{2}) = -1$.So the point is $(0, -1)$.The equation of the line is $y - (-1) = \frac{\sqrt{3}}{2}(x - 0)$,which simplifies to $2y + 2 = \sqrt{3}x$,or $\sqrt{3}x - 2y = 2$.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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