## Precalculus (6th Edition) Blitzer

$$y=\sqrt{3}x-16 \\ m=\sqrt{3}, \qquad b=-16$$
We find the line as follows: $$r \cos \left ( \theta + \frac{\pi }{6} \right )=8 \quad \Rightarrow \quad r \left ( \cos \theta \cos \frac{\pi }{6} - \sin \theta \sin \frac{\pi }{6} \right )=8 \quad \Rightarrow \quad \frac{\sqrt{3}}{2}r\cos \theta - \frac{1}{2} r \sin \theta =8 \quad \Rightarrow \quad \frac{\sqrt{3}}{2}x-\frac{1}{2}y=8 \quad \Rightarrow \quad y=\sqrt{3}x-16$$This line has slope $m=\sqrt{3}$ and $y$-intercept $b=-16$.