Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 9 Quadratic Relations and Conic Sections - Cumulative Review - Page 678: 35

Answer

$$\left(\frac{11}{2},\:6\right)$$

Work Step by Step

Using the distance formula, we find: $$\sqrt{\left(8-3\right)^2+\left(7-5\right)^2}\\ \sqrt{29}$$ Using the midpoint formula, we find: $$\left(\frac{8+3}{2},\:\frac{7+5}{2}\right)\\ \left(\frac{11}{2},\:6\right)$$
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