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 - Test: Chapter 10: 25

Answer

$\left( \dfrac{3}{2}, -6 \right)$

Work Step by Step

With the given points, then \begin{array}{l}\require{cancel} x_1= 2 ,\\x_2= 1 ,\\y_1= -5 ,\\y_2= -7 .\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{2+1}{2}, \dfrac{-5+(-7)}{2} \right) \\\\= \left( \dfrac{2+1}{2}, \dfrac{-5-7}{2} \right) \\\\= \left( \dfrac{3}{2}, \dfrac{-12}{2} \right) \\\\= \left( \dfrac{3}{2}, -6 \right) .\end{array}
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