A point in rectangular coordinates is given.Convert the point to polar coordinates,r 0. $( - \sqrt { 19 } , \sqrt { 19 } )$
A)Polar coordinates: $\left( \sqrt { 38 } , \frac { 3 \pi } { 4 } \right)$
B)Polar coordinates: $\left( \sqrt { 38 } , - \frac { \pi } { 4 } \right)$
C)Polar coordinates: $\left( \sqrt { 38 } , - \frac { 3 \pi } { 4 } \right)$
D)Polar coordinates: $\left( \sqrt { 38 } , \frac { \pi } { 4 } \right)$
E)Polar coordinates: $\left( \sqrt { 38 } , \frac { 5 \pi } { 4 } \right)$

Question

A point in rectangular coordinates is given.Convert the point to polar coordinates,r > 0.​ $( - \sqrt { 19 } , \sqrt { 19 } )$ ​
A)Polar coordinates: $\left( \sqrt { 38 } , \frac { 3 \pi } { 4 } \right)$
B)Polar coordinates: $\left( \sqrt { 38 } , - \frac { \pi } { 4 } \right)$
C)Polar coordinates: $\left( \sqrt { 38 } , - \frac { 3 \pi } { 4 } \right)$
D)Polar coordinates: $\left( \sqrt { 38 } , \frac { \pi } { 4 } \right)$
E)Polar coordinates: $\left( \sqrt { 38 } , \frac { 5 \pi } { 4 } \right)$

Quizplus · Accepted Answer

The magnitude $r = \sqrt{(-\sqrt{19})^2 + (\sqrt{19})^2} = \sqrt{38}$, and the angle $\theta = \tan^{-1}\left(\frac{\sqrt{19}}{-\sqrt{19}}\right) = \tan^{-1}(-1) = \frac{3\pi}{4}$ because the point is in the second quadrant.

A Point in Rectangular Coordinates Is Given (−19,19)( - \sqrt { 19 } , \sqrt { 19 } )(−19​,19​)

A Point in Rectangular Coordinates Is Given $( - \sqrt { 19 } , \sqrt { 19 } )$