College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Section 2.1 - Rectangular Coordinates and Graphs - 2.1 Exercises - Page 178: 16


$\text{a) Distance: } 2\sqrt{17} \text{ units}\\\text{b) Midpoint: } \left( 5,2 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Distance Formula and the Midpoint Formula to find the distance and the midpoint of the given points $\left( 6,-2 \right)$ and $\left( 4,6 \right).$ $\bf{\text{Solution Details:}}$ With the given points, then $x_1= 6 ,$ $x_2= 4 ,$ $y_1= -2 ,$ and $y_2= 6 .$ 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{(6-4)^2+(-2-6)^2} \\\\ d=\sqrt{(2)^2+(-8)^2} \\\\ d=\sqrt{4+64} \\\\ d=\sqrt{68} \\\\ d=\sqrt{4\cdot17} \\\\ d=\sqrt{(2)^2\cdot17} \\\\ d=2\sqrt{17} .\end{array} Using $\left( \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right)$ or the Midpoint Formula, then the midpoint of the line segment with endpoints given above is \begin{array}{l}\require{cancel} \left( \dfrac{6+4}{2}, \dfrac{-2+6}{2} \right) \\\\= \left( \dfrac{10}{2}, \dfrac{4}{2} \right) \\\\= \left( 5,2 \right) .\end{array} Hence, $ \text{a) Distance: } 2\sqrt{17} \text{ units}\\\text{b) Midpoint: } \left( 5,2 \right) .$
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