Answer
$\begin{array}{*{20}{c}}
{Point}\\
{}\\
A\\
B\\
C\\
D\\
E\\
F\\
G
\end{array}\begin{array}{*{20}{c}}
{Rectangular}\\
{\left( {x,y} \right)}\\
{\left( { - 3,3} \right)}\\
{\left( { - 3,0} \right)}\\
{\left( { - 2, - 1} \right)}\\
{\left( { - 1, - 1} \right)}\\
{\left( {1,1} \right)}\\
{\left( {2\sqrt 3 ,2} \right)}\\
{\left( {2\sqrt 3 , - 2} \right)}
\end{array}\begin{array}{*{20}{c}}
{Polar}\\
{\left( {r,\theta } \right)}\\
{\left( {3\sqrt 2 ,\frac{3}{4}\pi } \right)}\\
{\left( {3,\pi } \right)}\\
{\left( {\sqrt 5 ,\pi + {{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)} \right)}\\
{\left( {\sqrt 2 , \frac{5}{4}\pi } \right)}\\
{\left( {\sqrt 2 ,\frac{\pi }{4}} \right)}\\
{\left( {4,\frac{\pi }{6}} \right)}\\
{\left( {4, \frac{11\pi }{6}} \right)}
\end{array}$
Work Step by Step
Using the conversion formula from rectangular coordinates to polar coordinates given by
$r = \sqrt {{x^2} + {y^2}} $, ${\ \ \ }$ $\theta = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)$
we obtain the polar coordinates.
$\begin{array}{*{20}{c}}
{Point}\\
{}\\
A\\
B\\
C\\
D\\
E\\
F\\
G
\end{array}\begin{array}{*{20}{c}}
{Rectangular}\\
{\left( {x,y} \right)}\\
{\left( { - 3,3} \right)}\\
{\left( { - 3,0} \right)}\\
{\left( { - 2, - 1} \right)}\\
{\left( { - 1, - 1} \right)}\\
{\left( {1,1} \right)}\\
{\left( {2\sqrt 3 ,2} \right)}\\
{\left( {2\sqrt 3 , - 2} \right)}
\end{array}\begin{array}{*{20}{c}}
{Polar}\\
{\left( {r,\theta } \right)}\\
{\left( {3\sqrt 2 ,\frac{3}{4}\pi } \right)}\\
{\left( {3,\pi } \right)}\\
{\left( {\sqrt 5 ,\pi + {{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)} \right)}\\
{\left( {\sqrt 2 , \frac{5}{4}\pi } \right)}\\
{\left( {\sqrt 2 ,\frac{\pi }{4}} \right)}\\
{\left( {4,\frac{\pi }{6}} \right)}\\
{\left( {4, \frac{11\pi }{6}} \right)}
\end{array}$
