A point in rectangular coordinates is given.Convert the point to polar coordinates, r 0. $( - \sqrt { 3 } , - \sqrt { 3 } )$
A)Polar coordinates: $\left( \sqrt { 6 } , - \frac { \pi } { 4 } \right)$
B)Polar coordinates: $\left( \sqrt { 6 } , - \frac { 5 \pi } { 4 } \right)$
C)Polar coordinates: $\left( \sqrt { 6 } , \frac { 5 \pi } { 4 } \right)$
D)Polar coordinates: $\left( \sqrt { 6 } , \frac { \pi } { 4 } \right)$
E)Polar coordinates: $\left( \sqrt { 6 } , \frac { 3 \pi } { 4 } \right)$

Question

A point in rectangular coordinates is given.Convert the point to polar coordinates, r > 0.  $( - \sqrt { 3 } , - \sqrt { 3 } )$ 
A)Polar coordinates: $\left( \sqrt { 6 } , - \frac { \pi } { 4 } \right)$
B)Polar coordinates: $\left( \sqrt { 6 } , - \frac { 5 \pi } { 4 } \right)$
C)Polar coordinates: $\left( \sqrt { 6 } , \frac { 5 \pi } { 4 } \right)$
D)Polar coordinates: $\left( \sqrt { 6 } , \frac { \pi } { 4 } \right)$
E)Polar coordinates: $\left( \sqrt { 6 } , \frac { 3 \pi } { 4 } \right)$

Quizplus · Accepted Answer

The point $(- \sqrt{3}, - \sqrt{3})$ is in the third quadrant. The radius $r$ is $\sqrt{(-\sqrt{3})^2 + (-\sqrt{3})^2} = \sqrt{6}$. The angle $\theta$ is $\frac{5\pi}{4}$ in the third quadrant.

A Point in Rectangular Coordinates Is Given (−3,−3)( - \sqrt { 3 } , - \sqrt { 3 } )(−3​,−3​)

A Point in Rectangular Coordinates Is Given $( - \sqrt { 3 } , - \sqrt { 3 } )$