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 - Page 179: 33

Answer

$(5,-4)$

Work Step by Step

Let $(m,n)$ be the missing endpoint. Using $\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2} \right)$ or the Midpoint Formula, then the coordinates of the missing endpoint, given that the midpoint is $( 12,6 )$ and the other endpoint is $( 19,16 )$ are \begin{array}{l}\require{cancel} \dfrac{m+19}{2}=12 \\\text{AND}\\ \dfrac{n+16}{2}=6 .\end{array} Solving these equations separately results to \begin{array}{l}\require{cancel} m+19=2(12) \\\\ m+19=24 \\\\ m=24-19 \\\\ m=5 \\\text{AND}\\ n+16=2(6) \\\\ n+16=12 \\\\ n=12-16 \\\\ n=-4 .\end{array} Hence, the missing endpoint is $ (5,-4) .$
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