Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Secton 5.1 - The Unit Circle - 5.1 Exercises - Page 407: 13

Answer

$$P(\frac{3\sqrt 5}{7}, -\frac{2}{7})$$

Work Step by Step

Since $([], -\frac{2}{7})$ is an $(x, y)$ coordinate we can plug it into the equation of a circle, $x^{2}+y^{2}=1$. $$(x)^{2}+(-\frac{2}{7})^{2}=1$$ $$x^{2}+\frac{-2\times-2}{7\times7}=1$$ $$x^{2}+\frac{4}{49}=1$$ $$x^{2}=1-\frac{4}{49}$$ $$x^{2}=\frac{49}{49}-\frac{4}{49}$$ $$x^{2}=\frac{45}{49}$$ $$\sqrt (x^{2})=\sqrt (\frac{45}{49})$$ $$x=±\frac{3\sqrt 5}{7}$$ Since the coordinate needs to be in the 4nd quadrant, the x value needs to be positive. Therefore, $x=\frac{3\sqrt 5}{7}$ and $P(\frac{3\sqrt 5}{7}, -\frac{2}{7})$
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