# Geo5.4.10

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Find the equation of the line passing thro'the point $(2,\frac{\pi}{6})\,$ parallel and perpendicular to the line $\frac{3}{r}=4\cos\theta-3\sin\theta\,$

Given line is $\frac{3}{r}=4\cos\theta-3\sin\theta\,$

Any line parallel to this line is $\frac{k}{r}=4\cos\theta-3\sin\theta\,$

This line passes thro' $(2,\frac{\pi}{6})\,$

Therefore, $\frac{k}{2}=4\cos \frac{\pi}{6}-3\sin \frac{\pi}{6}\,$

$k=4\sqrt{3}-3\,$

Hence the line is $\frac{4\sqrt{3}-3}{r}=4\cos\theta-3\sin\theta\,$

Any line perpendicular to the given line is

$\frac{k_1}{r}=-4\sin\theta-3\cos\theta\,$

Given this line passes thro' $(2,\frac{\pi}{6})\,$

$\frac{k_1}{2}=-4\sin \frac{\pi}{6}-3\cos \frac{\pi}{6}\,$

$k_1=-(4+3\sqrt{3})\,$

The required line is $\frac{(4+3\sqrt{3})}{r}=4\sin\theta+3\cos\theta\,$

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