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: 128


$\sqrt{103} \text{ units}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Distance Formula to find the distance between the given points $\left( \sqrt{7},9\sqrt{3} \right)$ and $\left( -\sqrt{7},4\sqrt{3} \right)$. $\bf{\text{Solution Details:}}$ With the given points, then $x_1= \sqrt{7} ,$ $x_2= -\sqrt{7} ,$ $y_1= 9\sqrt{3} ,$ and $y_2= 4\sqrt{3} .$ 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{7}-(-\sqrt{7}))^2+(9\sqrt{3}-4\sqrt{3})^2} \\\\ d=\sqrt{(\sqrt{7}+\sqrt{7}))^2+(9\sqrt{3}-4\sqrt{3})^2} \\\\ d=\sqrt{(2\sqrt{7})^2+(5\sqrt{3})^2} \\\\ d=\sqrt{4(7)+25(3)} \\\\ d=\sqrt{28+75} \\\\ d=\sqrt{103} .\end{array} Hence, the distance is $ \sqrt{103} \text{ units} .$
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