## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 5 - Secton 5.1 - The Unit Circle - 5.1 Exercises: 9

#### Answer

$$P(-\frac{3}{5}, -\frac{4}{5})$$

#### Work Step by Step

Since ($-\frac{3}{5}, []$) is an (x, y) coordinate we can plug it into the equation of a circle, $x^{2}+y^{2}=1$. $$(-\frac{3}{5})^{2}+(y)^{2}=1$$ $$\frac{-3\times-3}{5\times5}+y^{2}=1$$ $$\frac{9}{25}+y^{2}=1$$ $$y^{2}=1-\frac{9}{25}$$ $$y^{2}=\frac{25}{25}-\frac{9}{25}$$ $$y^{2}=\frac{16}{25}$$ $$\sqrt (y^{2})=\sqrt (\frac{16}{25})$$ $$y=±\frac{4}{5}$$ Since the coordinate needs to be in the 3rd quadrant, both the x and y values need to be negative. Therefore, $y=-\frac{4}{5}$ and $$P(-\frac{3}{5}, -\frac{4}{5})$$

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