Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX B - Coordinate Geometry and Lines - B Exercises - Page A 15: 14

Answer

$a) |AB| + |BC| = |AC|$ $b) m_{1} = m_{2} = m{3}$

Work Step by Step

$a) |AB| = \sqrt (3+1)^2 + (11-3)^2 = \sqrt 80 = \sqrt (16 \times 5) = 4\sqrt 5$ $ |BC| = \sqrt (5-3)^2 + (15-11)^2 = \sqrt 20 = \sqrt (4 \times 5) = 2 \sqrt 5$ $|AC| = \sqrt (5+1)^2 + (15-3)^2 = \sqrt 180 = \sqrt (36 \times 5) = 6 \sqrt 5$ $|AB| + |BC| = |AC|$ $4 \sqrt 5 + 2 \sqrt 5 = 6 \sqrt 5$ $6 \sqrt 5 = 6 \sqrt 5$ $b) slope AB = m_{1} = \frac{11-3}{3+1} = 2$ $ slope BC = m_{2} = \frac{15-11}{5-3} = 2$ $ slope AC = m_{3} = \frac{15-3}{5+1} = 2$ $m_{1} = m_{2} = m_{3}$
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