Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - 2.4 Linear Functions - 2.4 Exercises - Page 232: 88

Answer

The distance between the first point and the fourth point is $3\sqrt{10}$ units.

Work Step by Step

RECALL: The distance $d$ between the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ Use the formula above to obtain: $d=\sqrt{(0-3)^2+(-6-3)^2} \\d=\sqrt{(-3)^2+(-9)^2} \\d=\sqrt{9+81} \\d=\sqrt{90} \\d=\sqrt{9(10)} \\d=3\sqrt{10}$
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