## Precalculus: Mathematics for Calculus, 7th Edition

$$P(-\frac{2}{3}, \frac{\sqrt 5}{3})$$
Since $(-\frac{2}{3}, [])$ is an $(x, y)$ coordinate we can plug it into the equation of a circle, $x^{2}+y^{2}=1$. $$(-\frac{2}{3})^{2}+(y)^{2}=1$$ $$\frac{-2\times-2}{3\times3}+y^{2}=1$$ $$\frac{4}{9}+y^{2}=1$$ $$y^{2}=1-\frac{4}{9}$$ $$y^{2}=\frac{9}{9}-\frac{4}{9}$$ $$y^{2}=\frac{5}{9}$$ $$\sqrt (y^{2})=\sqrt (\frac{5}{9})$$ $$y=±\frac{\sqrt 5}{3}$$ Since the coordinate needs to be in the 2nd quadrant, the y value has to be positive. Therefore, $y=\frac{\sqrt 5}{3}$ and $P(-\frac{2}{3}, \frac{\sqrt 5}{3})$