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: 37

Answer

The point $Q$ is closer to the point $R$

Work Step by Step

$P(3,1);$ $Q(-1,3);$ $R(-1,-1)$ 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 the points $P$ and $R$. For these two points, $x_{1}=3$, $y_{1}=1$, $x_{2}=-1$ and $y_{2}=-1$ Substitute these values into the formula: $d_{PR}=\sqrt{(-1-3)^{2}+(-1-1)^{2}}=\sqrt{(-4)^{2}+(-2)^{2}}=...$ $...=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\approx4.4721$ Find the distance between the points $Q$ and $R$. For these two points, $x_{1}=-1$, $y_{1}=3$, $x_{2}=-1$ and $y_{2}=-1$ Substitute these values into the formula: $d_{QR}=\sqrt{(-1+1)^{2}+(-1-3)^{2}}=\sqrt{0^{2}+(-4)^{2}}=...$ $...=\sqrt{0+16}=\sqrt{16}=4$ Since $d_{QR}\lt d_{PR}$, the point $Q$ is closer to the point $R$
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