Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Section 1.10 - Lines - 1.10 Exercises - Page 115: 81

Answer

$$m_{AD}=m_{BC}$$ $$m_{DC}=m_{AB}$$ $$m_{AD}m_{DC}=-1$$ $$m_{AB}m_{BC}=-1$$

Work Step by Step

We know, that each side of a rectangle has the same slope as its opposite side. Also, the sides connecting vertices are perpendicular (It means, that product of their slopes is $-1$). Above is visual representation of the points provided. We have to prove, that $m_{AD}=m_{BC}$ and $m_{DC}=m_{AB}$. To check whether or not its rectangle (each angle is $90°$) we also have to prove, that $m_{AD}m_{DC}=-1$ and $m_{AB}m_{BC}=-1$ $m_{AD}=\frac{6-1}{0-1}=\frac{5}{-1}=-5$ $m_{BC}=\frac{8-3}{10-11}=\frac{5}{-1}=-5$ $$m_{AD}=m_{BC}$$ $m_{DC}=\frac{8-6}{10-0}=\frac{2}{10}=\frac{1}{5}$ $m_{AB}=\frac{3-1}{11-1}=\frac{2}{10}=\frac{1}{5}$ $$m_{DC}=m_{AB}$$ We have found, that sides of this figure are parallel. $$m_{AD}m_{DC}=-5\times \frac{1}{5}=-1$$ $$m_{AB}m_{BC}=\frac{1}{5}\times (-5)=-1$$
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