Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter P - P.2 - Linear Models and Rates of Change - Exercises - Page 18: 83


Distance $=\dfrac{5\sqrt{2}}{2}$

Work Step by Step

Point: $(-2,1)$ $;$ Line: $x-y-2=0$ The distance between the point $(x_{1},y_{1})$ and the line $Ax+By+C=0$ is given by $\dfrac{|Ax_{1}+By_{1}+C|}{\sqrt{A^{2}+B^{2}}}$. In this case, $(x_{1},y_{1})=(-2,1)$ and $A=1$, $B=-1$ and $C=-2$ Substitute the known values into the formula and evaluate: Distance $=\dfrac{|(1)(-2)+(-1)(1)-2|}{\sqrt{1^{2}+(-1)^{2}}}=...$ $...=\dfrac{|-2-1-2|}{\sqrt{1+1}}=\dfrac{|-5|}{\sqrt{2}}=\dfrac{5}{\sqrt{2}}=...$ Rationalize the denominator: $...=\dfrac{5}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{5\sqrt{2}}{(\sqrt{2})^{2}}=\dfrac{5\sqrt{2}}{2}$
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