Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.3 - Simplifying Radicals, the Distance Formula, and Circles - 7.3 Exercises - Page 461: 127


$6\sqrt{2} \text{ units}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Distance Formula to find the distance between the given points $\left( \sqrt{2},\sqrt{6} \right)$ and $\left( -2\sqrt{2},4\sqrt{6} \right)$. $\bf{\text{Solution Details:}}$ With the given points, then $x_1= \sqrt{2} ,$ $x_2= -2\sqrt{2} ,$ $y_1= \sqrt{6} ,$ and $y_2= 4\sqrt{6} .$ Using the Distance Formula which is given by $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} ,$ then \begin{array}{l}\require{cancel} d=\sqrt{(\sqrt{2}-(-2\sqrt{2}))^2+(\sqrt{6}-4\sqrt{6})^2} \\\\ d=\sqrt{(\sqrt{2}+2\sqrt{2})^2+(\sqrt{6}-4\sqrt{6})^2} \\\\ d=\sqrt{(3\sqrt{2})^2+(-3\sqrt{6})^2} \\\\ d=\sqrt{9(2)+9(6)} \\\\ d=\sqrt{18+54} \\\\ d=\sqrt{72} \\\\ d=\sqrt{36\cdot2} \\\\ d=\sqrt{(6)^2\cdot2} \\\\ d=6\sqrt{2} .\end{array} Hence, the distance is $ 6\sqrt{2} \text{ units} .$
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.