Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.7 The Distance Formula, the Midpoint Formula, and Other Applications - 10.7 Exercise Set - Page 679: 59

Answer

$d\approx0.601 \text{ units} $

Work Step by Step

Using $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ or the Distance Formula, the distance, $d$, between the given points, $\left( \dfrac{1}{2}, \dfrac{1}{3} \right) \text{ and } \left( \dfrac{5}{6}, -\dfrac{1}{6} \right),$ is \begin{array}{l}\require{cancel} d=\sqrt{ \left(\dfrac{1}{2}-\dfrac{5}{6} \right)^2+\left(\dfrac{1}{3}-\left(-\dfrac{1}{6}\right) \right)^2} \\\\ d=\sqrt{ \left( -\dfrac{1}{3} \right)^2+\left(\dfrac{1}{2}\right)^2} \\\\ d=\sqrt{ \dfrac{1}{9}+\dfrac{1}{4}} \\\\ d=\sqrt{ \dfrac{13}{36}} \\\\ d= \dfrac{\sqrt{13}}{6} \\\\ d\approx0.601 \text{ units} .\end{array}
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