## Precalculus: Mathematics for Calculus, 7th Edition

$$P(-\frac{2\sqrt 2}{3}, \frac{1}{3})$$
Since $([], \frac{1}{3})$ is an $(x, y)$ coordinate we can plug it into the equation of a circle, $x^{2}+y^{2}=1$. $$(x)^{2}+(\frac{1}{3})^{2}=1$$ $$x^{2}+\frac{1\times1}{3\times3}=1$$ $$x^{2}+\frac{1}{9}=1$$ $$x^{2}=1-\frac{1}{9}$$ $$x^{2}=\frac{9}{9}-\frac{1}{9}$$ $$x^{2}=\frac{8}{9}$$ $$\sqrt (x^{2})=\sqrt (\frac{8}{9})$$ $$x=±\frac{2\sqrt 2}{3}$$ Since the coordinate needs to be in the 2nd quadrant, the x value needs to be negative. Therefore, $x=-\frac{2\sqrt 2}{3}$ and $P(-\frac{2\sqrt 2}{3}, \frac{1}{3})$