College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.1 - The Coordinate Plane - 1.1 Exercises - Page 92: 21

Answer

(a) $d=\sqrt{13}$ (b) midpoint=$(\frac{3}{2},1)$

Work Step by Step

(a) We find the distance between the points $(0,2)$ and $(3,0)$ by using the Distance Formula: $d=\sqrt{(3-0)^{2}+(0--2)^{2}}=\sqrt{3^{2}+2^{2}}=\sqrt{9+4}=\sqrt{13}$ (b) We find the midpoint by using the Midpoint Formula: midpoint=$(\displaystyle \frac{3+0}{2},\ \frac{0+2}{2})=(\frac{3}{2},1)$
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