Question

If the Cartesian coordinates of a point are $(-3,-\sqrt{3})$, then the polar coordinates are

• A. $(2\sqrt{3},\frac{5\pi }{6})$
• B. $(\sqrt{3},\frac{7\pi }{6})$
• C. $(2\sqrt{3},\frac{7\pi }{6})$
• D. $(2\sqrt{3},\frac{\pi }{6})$

Answer 1

The correct option is C $(2\sqrt{3},\frac{7\pi }{6})$

Given Cartesian coordinates are $(-3,-\sqrt{3})$.
We know that
$x=r\cos \theta ,y=r\sin \theta \Rightarrow r=\sqrt{x^2+y^2}\Rightarrow r=\sqrt{(-3{)}^2+(-\sqrt{3}{)}^2}=2\sqrt{3}$
And
$-3=2\sqrt{3}\cos \theta ,-\sqrt{3}=2\sqrt{3}\sin \theta \Rightarrow \cos \theta =-\frac{\sqrt{3}}{2},\sin \theta =-\frac{1}{2}$
As the point is in third quadrant,
$\Rightarrow \theta =\pi +\frac{\pi }{6}=\frac{7\pi }{6}$