Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.9 - Distance and Midpoint Formulas; Circles - Exercise Set - Page 280: 27

Answer

$\text{midpoint} = (1, 5\sqrt{5})$

Work Step by Step

RECALL: The midpoint of the line segment whose endpoints are $(x_1, y_1)$ and $(x_2, y_2)$ can be found using the midpoint formula: $\text{midpoint}=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right)$ Use the formula above to obtain: $\text{midpoint} = \left(\dfrac{8+(-6)}{2}, \dfrac{3\sqrt{5}+7\sqrt{5}}{2}\right)=\left(\dfrac{2}{2}, \dfrac{10\sqrt{5}}{2}\right) \\\text{midpoint} = (1, 5\sqrt{5})$
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