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

To convert from rectangular to polar coordinates, we use the formulas:

r = sqrt(x^2 + y^2)
theta = arctan(y/x)

Substituting the given values, we get:

r = sqrt(5^2 + 7^2) = sqrt(74)
theta = arctan(7/5)

Therefore, the polar coordinates are (sqrt(74), arctan(7/5)), which matches choice C.

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

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