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 16: 55

Answer

(a) The midpoint is $(4,9)$ (b) The midpoint is $(\frac{7}{2},-3)$

Work Step by Step

(a) We can find the midpoint of the line segment joining the two points $(1,3)$ and $(7,15)$: $x = \frac{1+7}{2} = 4$ $y = \frac{3+15}{2} = 9$ The midpoint is $(4,9)$ (b) We can find the midpoint of the line segment joining the two points $(-1,6)$ and $(8,-12)$: $x = \frac{-1+8}{2} = \frac{7}{2}$ $y = \frac{6+(-12)}{2} = -3$ The midpoint is $(\frac{7}{2},-3)$
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