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: 121


$9\sqrt{2} \text{ units}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Distance Formula to find the distance between the given points $\left( -6,5 \right)$ and $\left( 3,-4 \right)$. $\bf{\text{Solution Details:}}$ With the given points, then $x_1= -6 ,$ $x_2= 3 ,$ $y_1= 5 ,$ and $y_2= -4 .$ 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{(-6-3)^2+(5-(-4))^2} \\\\ d=\sqrt{(-6-3)^2+(5+4)^2} \\\\ d=\sqrt{(-9)^2+(9)^2} \\\\ d=\sqrt{81+81} \\\\ d=\sqrt{162} \\\\ d=\sqrt{81\cdot2} \\\\ d=\sqrt{(9)^2\cdot2} \\\\ d=9\sqrt{2} .\end{array} Hence, the distance is $ 9\sqrt{2} \text{ units} .$
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