Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 3 - Section 3.3 - Slope - Exercise Set - Page 242: 40

Answer

Opposite sides of the four sided figure lie on parallel lines, with one pair having slopes $-\displaystyle \frac{9}{5}$ and the other with $\displaystyle \frac{5}{9}.$

Work Step by Step

Use: $m=\displaystyle \frac{\text{change in y}}{\text{change in x}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$, parallel lines have equal slopes. Plot the points (see below). The slope of the line passing through $(-3,6)$ and $(6,11)$ is $m=\displaystyle \frac{11-6}{6-(-3)}=\frac{5}{9}$ The slope of the line passing through $(2,-3)$ and $(11,2)$ is $m=\displaystyle \frac{2-(-3)}{11-2}=\frac{5}{9}$ So, we have one pair of parallel lines passing through opposite sides. The slope of the line passing through $(-3,6)$ and $(2,-3)$ is $m=\displaystyle \frac{-3-6}{2-(-3)}=\frac{-9}{5}$ The slope of the line passing through $(6,11)$ and $(11,2)$ is $m=\displaystyle \frac{2-11}{11-6}=\frac{-9}{5}$, ... and we have another pair of parallel lines passing through opposite sides. Opposite sides of the four sided figure lie on parallel lines, with one pair having slopes $-\displaystyle \frac{9}{5}$ and the other with $\displaystyle \frac{5}{9}.$
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