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: 12

Answer

$\text{a) Distance: } 10 \text{ units}\\\text{b) Midpoint: } \left( -1, -1 \right)$

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 $\left( -4,3 \right)$ and $\left( 2,-5 \right).$ $\bf{\text{Solution Details:}}$ With the given points, then $x_1= -4 ,$ $x_2= 2 ,$ $y_1= 3 ,$ and $y_2= -5 .$ 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{(-4-2)^2+(3-(-5))^2} \\\\ d=\sqrt{(-4-2)^2+(3+5)^2} \\\\ d=\sqrt{(-6)^2+(8)^2} \\\\ d=\sqrt{36+64} \\\\ d=\sqrt{100} \\\\ d=\sqrt{(10)^2} \\\\ d=10 .\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{-4+2}{2}, \dfrac{3+(-5)}{2} \right) \\\\= \left( \dfrac{-4+2}{2}, \dfrac{3-5}{2} \right) \\\\= \left( \dfrac{-2}{2}, \dfrac{-2}{2} \right) \\\\= \left( -1, -1 \right) .\end{array} Hence, $ \text{a) Distance: } 10 \text{ units}\\\text{b) Midpoint: } \left( -1, -1 \right) .$
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