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: 18

Answer

$\text{a) Distance: } 3\sqrt{55} \text{ units}\\\text{b) Midpoint: } \left( 2\sqrt{7}, \dfrac{7\sqrt{3}}{2} \right)$

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 $\left( -\sqrt{7},8\sqrt{3} \right)$ and $\left( 5\sqrt{7}, -\sqrt{3} \right).$ $\bf{\text{Solution Details:}}$ With the given points, then $x_1= -\sqrt{7} ,$ $x_2= 5\sqrt{7} ,$ $y_1= 8\sqrt{3} ,$ and $y_2= -\sqrt{3} .$ 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{(-\sqrt{7}-5\sqrt{7})^2+(8\sqrt{3}-(-\sqrt{3}))^2} \\\\ d=\sqrt{(-\sqrt{7}-5\sqrt{7})^2+(8\sqrt{3}+\sqrt{3})^2} \\\\ d=\sqrt{(-6\sqrt{7})^2+(9\sqrt{3})^2} \\\\ d=\sqrt{36(7)+81(3)} \\\\ d=\sqrt{252+243} \\\\ d=\sqrt{495} \\\\ d=\sqrt{9\cdot55} \\\\ d=\sqrt{(3)^2\cdot55} \\\\ d=3\sqrt{55} .\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{-\sqrt{7}+5\sqrt{7}}{2}, \dfrac{8\sqrt{3}+(-\sqrt{3})}{2} \right) \\\\= \left( \dfrac{-\sqrt{7}+5\sqrt{7}}{2}, \dfrac{8\sqrt{3}-\sqrt{3}}{2} \right) \\\\= \left( \dfrac{4\sqrt{7}}{2}, \dfrac{7\sqrt{3}}{2} \right) \\\\= \left( 2\sqrt{7}, \dfrac{7\sqrt{3}}{2} \right) .\end{array} Hence, $ \text{a) Distance: } 3\sqrt{55} \text{ units}\\\text{b) Midpoint: } \left( 2\sqrt{7}, \dfrac{7\sqrt{3}}{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.