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