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: 130

Answer

$\sqrt{c^2-2cd+d^2+4d^2} \text{ units}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Distance Formula to find the distance between the given points $\left( c,c-d \right)$ and $\left( d,c+d \right)$. $\bf{\text{Solution Details:}}$ With the given points, then $x_1= c ,$ $x_2= d ,$ $y_1= c-d ,$ and $y_2= c+d .$ 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{(c-d)^2+(c-d-(c+d))^2} \\\\ d=\sqrt{(c-d)^2+(c-d-c-d)^2} \\\\ d=\sqrt{(c-d)^2+(-2d)^2} \\\\ d=\sqrt{(c-d)^2+4d^2} .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} d=\sqrt{c^2-2cd+d^2+4d^2} .\end{array} Hence, the distance is $ \sqrt{c^2-2cd+d^2+4d^2} \text{ units} .$
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