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 - Page 765: 30


$(3\sqrt2, 0)$

Work Step by Step

Simplify the radical coordinate of the first point to obtain: $(\sqrt{50}, -6) = (\sqrt{25(2)}, -6)=(5\sqrt2, -6)$ 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: $\text{midpoint} \\=\left(\dfrac{5\sqrt2+\sqrt2}{2}, \dfrac{-6+6}{2}\right) \\=\left(\dfrac{6\sqrt2}{2}, \dfrac{0}{2}\right) \\=(3\sqrt2, 0)$
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