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: 14

Answer

$\text{a) Distance: } \sqrt{202} \text{ units}\\\text{b) Midpoint: } \left( -\dfrac{5}{2}, -\dfrac{1}{2} \right)$

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 $\left( -8,4 \right)$ and $\left( 3,-5 \right).$ $\bf{\text{Solution Details:}}$ With the given points, then $x_1= -8 ,$ $x_2= 3 ,$ $y_1= 4 ,$ 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{(-8-3)^2+(4-(-5))^2} \\\\ d=\sqrt{(-8-3)^2+(4+5)^2} \\\\ d=\sqrt{(-11)^2+(9)^2} \\\\ d=\sqrt{121+81} \\\\ d=\sqrt{202} .\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{-8+3}{2}, \dfrac{4+(-5)}{2} \right) \\\\= \left( \dfrac{-8+3}{2}, \dfrac{4-5}{2} \right) \\\\= \left( \dfrac{-5}{2}, \dfrac{-1}{2} \right) \\\\= \left( -\dfrac{5}{2}, -\dfrac{1}{2} \right) .\end{array} Hence, $ \text{a) Distance: } \sqrt{202} \text{ units}\\\text{b) Midpoint: } \left( -\dfrac{5}{2}, -\dfrac{1}{2} \right) .$
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