The polar co-ordinates of the point, whose Cartesian co-ordinates are $(-2 \sqrt{3}, 2)$, are

The polar co-ordinates of the point, whose Cartesian co-ordinates are $(-2 \sqrt{3}, 2)$, are
  1. $\left(4,\left(\frac{11 \pi}{12}\right)\right)$
  2. $\left(4,\left(\frac{5 \pi}{6}\right)\right)$
  3. $\left(4,\left(\frac{3 \pi}{4}\right)\right)$
  4. $\left(4,\left(\frac{2 \pi}{3}\right)\right)$

Solution

$\begin{aligned} & x=-2 \sqrt{3} \text { and } y=2 \\ & r=\sqrt{x^2+y^2}=\sqrt{12+4}=4 \\ & \theta=\pi-\tan ^{-1}\left|\frac{2}{-2 \sqrt{3}}\right|=\pi-\tan ^{-1} \frac{1}{\sqrt{3}}=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}\end{aligned}$

Asked in: MHT CET 2022 (06 Aug Shift 2)

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