Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.9 - The Coordinate Plane; Graphs of Equations; Circles - 1.9 Exercises - Page 102: 21

Answer

distance between the two points = $\sqrt{13}$ midpoint of the segment that joins the points: $(1.5, 1)$

Work Step by Step

RECALL: (i) The distance, d, between the points (a, b) and (c, d) can be found using the distance formula: $\\d=\sqrt{(a-c)^2+(b-d)^2}$ (ii) The midpoint of the points (a, b) and (c, d) can be found using the midpoint formula: $\\\text{midpoint} = \left(\frac{a+c}{2}, \frac{b+d}{2}\right)$ The two points are (0, 2) and (3, 0). Use the formulas above to have: (a) distance between the two points: $=\sqrt{(0-3)^2+(2-0)^2} \\=\sqrt{9+4} \\=\sqrt{13}$ (b) midpoint of the segment that joins them: $\\=\left(\frac{0+3}{2}, \frac{2+0}{2}\right) \\=(1.5, 1)$
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