Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.3 - Polar Coordinates - Concept and Vocabulary Check - Page 741: 3


The given point lies in quadrant II.

Work Step by Step

Here, $\left( r,\theta \right)$ is a polar coordinate defined by the formula as: $p\left( r,\theta \right)=r\cos \theta +ir\sin \theta $ …….(1) Define the real value $r\cos \theta$ on the x-axis and define the imaginary value $r\sin \theta $ on the y-axis. Then, put the value of $\left( r,\theta \right)=\left( 4,135{}^\circ \right)$ in equation (1), Therefore, $\begin{align} & p\left( r,\theta \right)=r\cos \theta +ir\sin \theta \\ & =\left( 4\cos 135{}^\circ \right)+i\left( 4\sin 135{}^\circ \right) \\ & =\left( 4\times \left( -0.707 \right) \right)+i\left( 4\times \left( 0.0883 \right) \right) \\ & =\left( -2.828 \right)+i\left( 0.3532 \right) \end{align}$ Now, by comparing it with the equation in the rectangular coordinates, we get: $p\left( x,y \right)=x+iy$ Hence, the rectangular coordinate is: $x=-2.828\,\,\,\text{ and }\,\,y=0.3532$ So, $\left( x,y \right)=\left( -2.828,0.3532 \right)$ Since $x=-2.828$ is negative in nature then it will lie in the second and third quadrant, while $y=0.3532$ is positive so it will lie in the first and second quadrant. So the point lies in quadrant II.
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