Answer
Rectangular coordinates are $\left( -6,0 \right)$, and $\left( \frac{5\sqrt{2}}{2},-\frac{5\sqrt{2}}{2} \right)$ ; Distance is $\sqrt{61+30\sqrt{2}}$
Work Step by Step
We will use $x=r\cos \theta $ and $y=r\sin \theta $ to determine rectangular coordinates.
For, $\left( 6,\pi \right)$
$\begin{align}
& x=6\cos \pi \\
& =6\left( -1 \right) \\
& =-6
\end{align}$
$\begin{align}
& y=6sin\pi \\
& =6\left( 0 \right) \\
& =0
\end{align}$
For, $\left( 5,\frac{7\pi }{4} \right)$
$\begin{align}
& x=5\cos \frac{7\pi }{4} \\
& =5\left( \frac{1}{\sqrt{2}} \right) \\
& =\frac{5}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} \\
& =\frac{5\sqrt{2}}{2}
\end{align}$
$\begin{align}
& y=5sin\frac{7\pi }{4} \\
& =5\left( -\frac{1}{\sqrt{2}} \right) \\
& =-\frac{5}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} \\
& =-\frac{5\sqrt{2}}{2}
\end{align}$
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,
$\begin{align}
& \text{D}=\sqrt{{{\left( 0-\left( -\frac{5\sqrt{2}}{2} \right) \right)}^{2}}+{{\left( -6-\frac{5\sqrt{2}}{2} \right)}^{2}}} \\
& =\sqrt{{{\left( \frac{5\sqrt{2}}{2} \right)}^{2}}+{{\left( -6-\frac{5\sqrt{2}}{2} \right)}^{2}}} \\
& =\sqrt{\frac{25}{2}+36+\frac{25}{2}+30\sqrt{2}} \\
& =\sqrt{61+30\sqrt{2}}
\end{align}$
Therefore, rectangular coordinate for $\left( 6,\pi \right)$ is $\left( -6,0 \right)$ and for $\left( 5,\frac{7\pi }{4} \right)$ is $\left( \frac{5\sqrt{2}}{2},-\frac{5\sqrt{2}}{2} \right)$ and the distance between these points is $\sqrt{61+30\sqrt{2}}$.
