College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Section 2.1 - Rectangular Coordinates and Graphs - 2.1 Exercises: 17

Answer

(a) $d=\sqrt{133}$ (b) $midpoint\displaystyle =(2\sqrt{2}, \frac{3\sqrt{5}}{2})$

Work Step by Step

We are given the points: $P(3\sqrt{2},4\sqrt{5})$ , $Q(\sqrt{2},-\sqrt{5})$ (a) To find the distance between the two points, we use the distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ $=\sqrt{(\sqrt{2}-3\sqrt{2})^{2}+(-\sqrt{5}-4\sqrt{5})^{2}}$ $=\sqrt{(-2\sqrt{2})^{2}+(-5\sqrt{5})^{2}}$ $=\sqrt{8+125}$ $=\sqrt{133}$ (b) To find the midpoint, we use the midpoint formula: $\displaystyle Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ $\displaystyle =(\frac{3\sqrt{2}+\sqrt{2}}{2},\frac{4\sqrt{5}+-\sqrt{5}}{2})$ $\displaystyle =(\frac{4\sqrt{2}}{2},\frac{3\sqrt{5}}{2})$ $\displaystyle =(2\sqrt{2}, \frac{3\sqrt{5}}{2})$
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