## Precalculus (6th Edition) Blitzer

The polar coordinates of $\left( 0,-6 \right)$ are $\left( 6,\frac{3\pi }{2} \right)$.
The polar coordinates of the point are $\left( r,\theta \right)$. Now, rewrite the polar coordinates in terms of rectangular coordinates as below: $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ …… (1) $\tan \theta =\frac{y}{x}$ …… (2) Substituting the values of $x\ \text{ and }\ y$ in (1) and (2), we get \begin{align} & r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\ & =\sqrt{{{\left( 0 \right)}^{2}}+{{\left( -6 \right)}^{2}}} \\ & =\sqrt{36} \\ & r=6 \end{align} And, \begin{align} & \tan \theta =\frac{y}{x} \\ & =\frac{-6}{0} \\ & \tan \theta =\infty \left( \text{undefined} \right) \end{align} Hence, $\tan \theta =\infty$ And, $\tan \frac{\pi }{2}=\infty$ The $\theta$ lies on the negative y-axis which means \begin{align} & \theta =\pi +\frac{\pi }{2} \\ & =\frac{3\pi }{2} \end{align} Therefore, the polar coordinates of $\left( 0,-6 \right)$ are $\left( 6,\frac{3\pi }{2} \right)$.