Question

The polar coordinates of the point whose Cartesian coordinates are $(-1,-\sqrt{3})$, are

• A. $(2,\frac{\pi }{3})$
• B. $(2,-\frac{2\pi }{3})$
• C. $(2,\frac{2\pi }{3})$
• D. $(4,-\frac{2\pi }{3})$

Answer 1

The correct option is B $(2,-\frac{2\pi }{3})$

$(x,y)\equiv (-1,-\sqrt{3})$
Since, $r=\sqrt{x^2+y^2}$
$\Rightarrow r=\sqrt{(-1{)}^2+(-\sqrt{3}{)}^2}$
$\Rightarrow r=\sqrt{1+3}=2$

Also, $(x,y)\equiv (-1,-\sqrt{3})$ lies in $3^{\text{rd}}$ quadrant.
$\therefore \theta =-\pi +{\tan }^{-1}\left|\frac{y}{x}\right|$
$\Rightarrow \theta =-\pi +{\tan }^{-1}\left|\frac{-\sqrt{3}}{-1}\right|$
$\Rightarrow \theta =-\pi +{\tan }^{-1}\left(\sqrt{3}\right)$
$\Rightarrow \theta =-\pi +\frac{\pi }{3}=-\frac{2\pi }{3}$
$\therefore (r,\theta )\equiv (2,-\frac{2\pi }{3})$