## Precalculus (6th Edition) Blitzer

Rectangular coordinates are $\left( -6,0 \right)$, and $\left( \frac{5\sqrt{2}}{2},-\frac{5\sqrt{2}}{2} \right)$ ; Distance is $\sqrt{61+30\sqrt{2}}$
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 $\sqrt{61+30\sqrt{2}}$.