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: 15

Answer

$\text{a) Distance: } 3\sqrt{41} \text{ units}\\\text{b) Midpoint: } \left( 0, \dfrac{5}{2} \right)$

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 $\left( -6,-5 \right)$ and $\left( 6,10 \right).$ $\bf{\text{Solution Details:}}$ With the given points, then $x_1= -6 ,$ $x_2= 6 ,$ $y_1= -5 ,$ and $y_2= 10 .$ 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{(-6-6)^2+(-5-10)^2} \\\\ d=\sqrt{(-12)^2+(-15)^2} \\\\ d=\sqrt{144+225} \\\\ d=\sqrt{369} \\\\ d=\sqrt{9\cdot41} \\\\ d=\sqrt{(3)^2\cdot41} \\\\ d=3\sqrt{41} .\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+6}{2}, \dfrac{-5+10}{2} \right) \\\\= \left( \dfrac{0}{2}, \dfrac{5}{2} \right) \\\\= \left( 0, \dfrac{5}{2} \right) .\end{array} Hence, $ \text{a) Distance: } 3\sqrt{41} \text{ units}\\\text{b) Midpoint: } \left( 0, \dfrac{5}{2} \right) .$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.