Precalculus (6th Edition)

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

Chapter 2 - Graphs and Functions - 2.1 Rectangular Coordinates and Graphs - 2.1 Exercises - Page 194: 46

Answer

$d=[(x_2-x_1)^2+(y_2-y_1)^2]^{1/2}$

Work Step by Step

Since $\sqrt x=x^{1/2}$, we can rewrite the distance formula as $d=[(x_2-x_1)^2+(y_2-y_1)^2]^{1/2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.