College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Section 2.4 - Linear Functions - 2.4 Exercises - Page 215: 85

Answer

Midpoint = $(3,3)$ This matches the middle entry in the table.

Work Step by Step

To find the midpoint between points $(x_1,y_1)$ and $(x_2,y_2)$, we use the midpoint formula: $\displaystyle Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ Using points $(0,-6)$ and $(6,12)$, we get: $Midpoint=(\displaystyle \frac{0+6}{2},\frac{-6+12}{2})$ =$(6/2,6/2)$ =$(3,3)$ The middle entry in the table is $(3,3)$, the same as the midpoint (as we would expect).
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