Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 1 - Graphs - Chapter Review - Review Exercises - Page 41: 3

Answer

a) $12.$ b)$(4,2)$.

Work Step by Step

The distance formula from $P_1(x_1,y_1)$ to $P_2(x_2,y_2)$ is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. The midpoint $M$ of the line segment from $P_1(x_1,y_1)$ to $P_2(x_2,y_2)$ is: $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$. Hence: a) $d=\sqrt{(4-4)^2+(-4-8)^2}=\sqrt{0+144}=\sqrt{144}=12.$ b)$M=(\frac{4+4}{2},\frac{-4+8}{2})=(4,2)$.
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