Convert a Point from Polar to Rectangular Coordinates
-$\left( 9 , \frac { 3 \pi } { 4 } \right)$
A) $\left( \frac { - 9 \sqrt { 2 } } { 2 } , \frac { 9 \sqrt { 2 } } { 2 } \right)$
B) $\left( \frac { 9 \sqrt { 2 } } { 2 } , \frac { - 9 \sqrt { 2 } } { 2 } \right)$
C) $\left( \frac { 9 \sqrt { 2 } } { 2 } , \frac { 9 \sqrt { 2 } } { 2 } \right)$
D) $\left( \frac { - 9 \sqrt { 2 } } { 2 } , \frac { - 9 \sqrt { 2 } } { 2 } \right)$

Question

Quizplus · Accepted Answer

To convert polar coordinates $(r, \theta)$ to rectangular coordinates $(x, y)$, use the formulas $x = r \cos(\theta)$ and $y = r \sin(\theta)$. For $\left( 9 , \frac { 3 \pi } { 4 } \right)$, $\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$ and $\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}$. Thus, $x = 9 \cdot -\frac{\sqrt{2}}{2} = \frac{-9\sqrt{2}}{2}$ and $y = 9 \cdot \frac{\sqrt{2}}{2} = \frac{9\sqrt{2}}{2}$, making the correct answer $\left( \frac { - 9 \sqrt { 2 } } { 2 } , \frac { 9 \sqrt { 2 } } { 2 } \right)$.

Convert a Point from Polar to Rectangular Coordinates - (9,3π4)\left( 9 , \frac { 3 \pi } { 4 } \right)(9,43π​)

Convert a Point from Polar to Rectangular Coordinates
- $\left( 9 , \frac { 3 \pi } { 4 } \right)$