Answer
$\begin{array}{*{20}{c}}
{Rectangular}&{Polar}\\
{\left( {x,y} \right)}&{\left( {r,\theta } \right)}\\
{\left( {1, - 3} \right)}&{\left( {\sqrt {10} ,5.034} \right)}\\
{\left( {3, - 1} \right)}&{\left( {\sqrt {10} ,5.961} \right)}
\end{array}$
Work Step by Step
We have $\left( {x,y} \right) = \left( {1, - 3} \right)$.
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:
$r = \sqrt {1 + {{\left( { - 3} \right)}^2}} = \sqrt {10} $,
$\theta = {\tan ^{ - 1}}\left( {\frac{{ - 3}}{1}} \right) \simeq - 1.249$ ${\ \ }$ or ${\ \ }$ $\theta \simeq 2\pi - 1.249 \simeq 5.034$
$\left( {r,\theta } \right) = \left( {\sqrt {10} ,5.034} \right)$
Similarly for the point $\left( {x,y} \right) = \left( {3, - 1} \right)$.
$r = \sqrt {{3^2} + {{\left( { - 1} \right)}^2}} = \sqrt {10} $,
$\theta = {\tan ^{ - 1}}\left( {\frac{{ - 1}}{3}} \right) \simeq - 0.322$ ${\ \ }$ or ${\ \ }$ $\theta \simeq 2\pi - 0.322 \simeq 5.961$
$\left( {r,\theta } \right) = \left( {\sqrt {10} ,5.961} \right)$
In summary:
$\begin{array}{*{20}{c}}
{Rectangular}&{Polar}\\
{\left( {x,y} \right)}&{\left( {r,\theta } \right)}\\
{\left( {1, - 3} \right)}&{\left( {\sqrt {10} ,5.034} \right)}\\
{\left( {3, - 1} \right)}&{\left( {\sqrt {10} ,5.961} \right)}
\end{array}$
