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: 36

Answer

The point $C$ is closer to the point $E$

Work Step by Step

$C(-6,3);$ $D(3,0);$ $E(-2,1)$ The distance between two points is given by the formula $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ Find the distance between points $C$ and $E$. For these two points, $x_{1}=-6$, $y_{1}=3$, $x_{2}=-2$ and $y_{2}=1$ Substitute these values into the formula: $d_{CE}=\sqrt{(-2+6)^{2}+(1-3)^{2}}=\sqrt{4^{2}+(-2)^{2}}=...$ $...=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\approx4.4721$ Find the distance between the points $D$ and $E$. For these two points, $x_{1}=3$, $y_{1}=0$, $x_{2}=-2$ and $y_{2}=1$ Substitute these values into the formula: $d_{DE}=\sqrt{(-2-3)^{2}+(1-0)^{2}}=\sqrt{(-5)^{2}+1^{2}}=...$ $...=\sqrt{25+1}=\sqrt{26}\approx5.099$ Since $d_{CE}\lt d_{DE}$, the point $C$ is closer to the point $E$
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