Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 1 - Section 1.2 - The Rectangular Coordinate System - 1.2 Problem Set - Page 25: 54


$(-0.7071,0.7071), (-0.7071,-0.7071)$

Work Step by Step

The line $x=-\frac{\sqrt{2}}{2}$ intersects the circle $x^2+y^2=1$ at $(-0.7071,0.7071)\,$and$\,(-0.7071,-0.7071)$ as can be seen from the figure. The solution can also be obtained analytically by substituting $x=-\frac{\sqrt{2}}{2}$ in the equation $x^2+y^2=1$ and solving for $y$ $$y^2 = 1-x^2$$ $$y = \pm \sqrt{1-x^2} = \pm \sqrt{1-(-\frac{\sqrt{2}}{2})^2} = \pm \sqrt{0.5}$$ $$\therefore y = \pm \frac{\sqrt{2}}{2} \approx \pm 0.7071$$ $\therefore $ The coordinates are $(-0.7071,0.7071)\,$and$\,(-0.7071,-0.7071)$.
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