Answer
The point has coordinates $(\dfrac{3 \sqrt 2}{2},\dfrac{ 3\sqrt 2}{2} )$ 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 $r=- 3$ and $ \theta= \dfrac{- 3\pi}{3}$
Plug these values in equation (1) to obtain:
$x=(-3) \cos(\dfrac{- 3\pi}{4})=(-3)(\dfrac{-\sqrt 2}{2})=\dfrac{3 \sqrt 2}{2} \\
y=(-3) \sin(\dfrac{- 3\pi}{4})=(-3)(\dfrac{-\sqrt 2}{2})=\dfrac{ 3 \sqrt 2}{2}$
Therefore, the point has coordinates $(\dfrac{3 \sqrt 2}{2},\dfrac{ 3\sqrt 2}{2} )$ in the rectangular coordinate system.
