Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 10 - Section 10.1 - Distance and Midpoint Formulas; Circles - Exercise Set: 26

Answer

$\left(-\dfrac{2}{5}, \dfrac{1}{10}\right)$

Work Step by Step

RECALL: The midpoint of $(x_1, y_1)$ and $(x_2, y_2)$ can be found using the midpoint formula $\text{midpoint}=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ Use the formula above to obtain: $\text{midpoint} \\=\left(\dfrac{-\frac{2}{5}+(-\frac{2}{5})}{2}, \dfrac{\frac{7}{15}+(-\frac{4}{15})}{2}\right) \\=\left(\dfrac{-\frac{4}{5}}{2}, \frac{\frac{3}{15}}{2}\right) \\=\left(-\dfrac{4}{5(2)}, \dfrac{3}{15(2)}\right) \\=\left(-\dfrac{4}{10}, \dfrac{3}{30}\right) \\=\left(-\dfrac{2}{5}, \dfrac{1}{10}\right)$
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