College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.1 - The Coordinate Plane - 1.1 Exercises - Page 92: 37


Point $Q(-1,3)$ is closer to point $R$ .

Work Step by Step

$d(P, R)=\sqrt{(-1-3)^{2}+(-1-1)^{2}}=\sqrt{(-4)^{2}+(-2)^{2}}$ $=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$. $d(Q, R)=\sqrt{(-1-(-1))^{2}+(-1-3)^{2}}$ $=\sqrt{0+(-4)^{2}}=\sqrt{16}=4$. $4<2\sqrt{5}$, so point $Q(-1,3)$ is closer to point $R$ than point P.
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