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: 38

Answer

$ a.\qquad$ The distances are equal ($\sqrt{58}$) $ b.\qquad$ The distances are equal ($\sqrt{a^{2}+b^{2}}$)

Work Step by Step

a. The distance from (7, 3) to the origin is $\sqrt{(7-0)^{2}+(3-0)^{2}}=\sqrt{7^{2}+3^{2}}=\sqrt{49+9}=\sqrt{58}$. The distance from (3, 7) to the origin is $\sqrt{(3-0)^{2}+(7-0)^{2}}=\sqrt{3^{2}+7^{2}}=\sqrt{9+49}=\sqrt{58}$. So, the points are the same distance from the origin. b. The distance from $(a, b)$ to the origin is $\sqrt{(a-0)^{2}+(b-0)^{2}}=\sqrt{a^{2}+b^{2}}$. The distance from $(b, a)$ to the origin is $\sqrt{(b-0)^{2}+(a-0)^{2}}=\sqrt{b^{2}+a^{2}}=\sqrt{a^{2}+b^{2}}$. So, the points are the same distance from the origin.
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