Answer
Rectangular coordinates are $\left( -1,\sqrt{3} \right)$, $\left( 2\sqrt{3},2 \right)$; Distance is $2\sqrt{5}$
Work Step by Step
Using $x=r\cos \theta $ and $y=r\sin \theta $ we will determine rectangular coordinates.
For, $\left( 2,\frac{2\pi }{3} \right)$
$\begin{align}
& x=2\cos \frac{2\pi }{3} \\
& =2\left( -\frac{1}{2} \right) \\
& =-1
\end{align}$
$\begin{align}
& y=2sin\frac{2\pi }{3} \\
& =2\left( \frac{\sqrt{3}}{2} \right) \\
& =\sqrt{3}
\end{align}$
For, $\left( 4,\frac{\pi }{6} \right)$
$\begin{align}
& x=4\cos \frac{\pi }{6} \\
& =4\left( \frac{\sqrt{3}}{2} \right) \\
& =2\sqrt{3}
\end{align}$
$\begin{align}
& y=4sin\frac{\pi }{6} \\
& =4\left( \frac{1}{2} \right) \\
& =2
\end{align}$
Using, $\sqrt{{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}}$ we will determine the distance between two points as,
$\begin{align}
& \text{D}=\sqrt{{{\left( 2-\sqrt{3} \right)}^{2}}+{{\left( 2\sqrt{3}-\left( -1 \right) \right)}^{2}}} \\
& =\sqrt{{{\left( 2-\sqrt{3} \right)}^{2}}+{{\left( 2\sqrt{3}+1 \right)}^{2}}} \\
& =\sqrt{4-4\sqrt{3}+3+12+4\sqrt{3}+1} \\
& =\sqrt{20}
\end{align}$
This implies that:
$\text{D}=2\sqrt{5}$
Therefore, rectangular coordinate for $\left( 2,\frac{2\pi }{3} \right)$ is $\left( -1,\sqrt{3} \right)$ and for $\left( 4,\frac{\pi }{6} \right)$ is $\left( 2\sqrt{3},2 \right)$ and the distance between these points is $2\sqrt{5}$.
