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: 11

Answer

$\text{a) Distance: } 8\sqrt{2} \\\text{b) Midpoint: } \left( -9, -3 \right)$

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 $\left( -5,-7 \right)$ and $\left( -13,1 \right).$ $\bf{\text{Solution Details:}}$ With the given points, then $x_1= -5 ,$ $x_2= -13 ,$ $y_1= -7 ,$ and $y_2= 1 .$ 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{(-5-(-13))^2+(-7-1)^2} \\\\ d=\sqrt{(-5+13)^2+(-7-1)^2} \\\\ d=\sqrt{(8)^2+(-8)^2} \\\\ d=\sqrt{64+64} \\\\ d=\sqrt{128} \\\\ d=\sqrt{64\cdot2} \\\\ d=\sqrt{(8)^2\cdot2} \\\\ d=8\sqrt{2} .\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{-5+(-13)}{2}, \dfrac{-7+1}{2} \right) \\\\= \left( \dfrac{-5-13}{2}, \dfrac{-7+1}{2} \right) \\\\= \left( \dfrac{-18}{2}, \dfrac{-6}{2} \right) \\\\= \left( -9, -3 \right) .\end{array} Hence, $ \text{a) Distance: } 8\sqrt{2} \\\text{b) Midpoint: } \left( -9, -3 \right) .$
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