Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - Review Exercises: Chapter 10: 51

Answer

$\left( 2,-\dfrac{3}{2} \right)$

Work Step by Step

With the given points, then \begin{array}{l}\require{cancel} x_1= -7 ,\\x_2= 3 ,\\y_1= -2 ,\\y_2= -1 .\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 the endpoints given above is \begin{array}{l}\require{cancel} \left( \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right) \\\\= \left( \dfrac{-7+3}{2}, \dfrac{-2+(-1)}{2} \right) \\\\= \left( \dfrac{-7+3}{2}, \dfrac{-2-1}{2} \right) \\\\= \left( \dfrac{4}{2}, \dfrac{-3}{2} \right) \\\\= \left( 2,-\dfrac{3}{2} \right) .\end{array}
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