Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.9 - The Coordinate Plane; Graphs of Equations; Circles - 1.9 Exercises - Page 102: 38

Answer

$a)$ Points $(7,3)$ and $(3,7)$ are the same distance from the origin. $b)$ Points $(a,b)$ and $(b,a)$ are the same distance from the origin.

Work Step by Step

Use the formula for the distance between two points to do this problem. The formula is $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ $a)$ $(7,3)$ $;$ $(3,7)$ Find the distance between point $(7,3)$ and the origin: $d_{1}=\sqrt{(7-0)^{2}+(3-0)^{2}}=\sqrt{49+9}=\sqrt{50}=5\sqrt{2}$ Find the distance between point $(3,7)$ and the origin: $d_{2}=\sqrt{(3-0)^{2}+(7-0)^{2}}=\sqrt{9+49}=\sqrt{50}=5\sqrt{2}$ Since $d_{1}=d_{2}$, points $(7,3)$ and $(3,7)$ are the same distance from the origin. $b)$ $(a,b)$ $;$ $(b,a)$ Find the distance between point $(a,b)$ and the origin: $d_{1}=\sqrt{(a-0)^{2}+(b-0)^{2}}=\sqrt{a^{2}+b^{2}}$ Find the distance between point $(b,a)$ and the origin: $d_{2}=\sqrt{(b-0)^{2}+(a-0)^{2}}=\sqrt{b^{2}+a^{2}}=\sqrt{a^{2}+b^{2}}$ Since $d_{1}=d_{2}$, points $(a,b)$ and $(b,a)$ are the same distance from the origin.
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