# Geo5.4.12

Find the equation of the line passing thro' $(-1,\frac{\pi}{2})\,$ and parallel to $4=r(2\cos\theta+\sqrt{3}\sin\theta)\,$

Given line is $\frac{4}{r}=2\cos\theta+\sqrt{3}\sin\theta\,$

Any line parallel to the line is

$\frac{k}{r}=2\cos\theta+\sqrt{3}\sin\theta\,$

Given this line passes thro' $(-1,\frac{\pi}{2})\,$

Hence

$\frac{k}{-1}=2\cos\frac{\pi}{2}+\sqrt{3}\sin\frac{\pi}{2}\,$

$-k=2(0)+\sqrt{3}\,$

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

Therefore the required line is $-\sqrt{3}=r(2\cos\theta+\sqrt{3}\sin\theta\,$

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