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: 129


$\sqrt{5y^2-2xy+x^2} \text{ units}$

Work Step by Step

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