## Precalculus (6th Edition) Blitzer

Rectangular coordinates are $\left( -1,\sqrt{3} \right)$, $\left( 2\sqrt{3},2 \right)$; Distance is $2\sqrt{5}$
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 $2\sqrt{5}$.