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: 60

Answer

$d\approx0.915 \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, $\left( \dfrac{5}{7}, \dfrac{1}{14} \right) \text{ and } \left( \dfrac{1}{7}, \dfrac{11}{14} \right),$ is \begin{array}{l}\require{cancel} d=\sqrt{ \left(\dfrac{5}{7}-\dfrac{1}{7} \right)^2+\left(\dfrac{1}{14}-\dfrac{11}{14} \right)^2} \\\\ d=\sqrt{ \left( \dfrac{4}{7} \right)^2+\left(\dfrac{-10}{14}\right)^2} \\\\ d=\sqrt{ \dfrac{16}{49}+\dfrac{100}{196}} \\\\ d=\sqrt{ \dfrac{41}{49}} \\\\ d= \dfrac{\sqrt{41}}{7} \\\\ d\approx0.915 \text{ units} .\end{array}
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