Answer
The point $(x,y)$ has coordinates $(-\dfrac{1}{2} ,\dfrac{\sqrt 3}{2} )$ in the rectangular coordinate system.
Work Step by Step
The point $(x,y)$ in the rectangular coordinate system can be expressed as:
$x=r \ \cos(\theta)$, $y=r \ \sin(\theta) ...(1)$
Here, we have $r=-1$ and $ \theta= \dfrac{-\pi}{3}$
Plug these values in equation (1) to obtain:
$x=(-1) \cos(\dfrac{-\pi}{3})=(-1)(\dfrac{1}{2})=-\dfrac{1}{2} \\
y=(-1) \sin(\dfrac{-\pi}{3})=(-1)(\dfrac{-\sqrt 3}{2})=\dfrac{\sqrt 3}{2}$
Therefore, the point has coordinates $(-\dfrac{1}{2} ,\dfrac{\sqrt 3}{2} )$ in the rectangular coordinate system.
