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: 85


Distance $=2\sqrt{2}$

Work Step by Step

Line: $x+y=1$ $;$ Line: $x+y=5$ The distance between the point $(x_{1},y_{1})$ and the line $Ax+By+C=0$ is given by $\dfrac{|Ax_{1}+By_{1}+C|}{\sqrt{A^{2}+B^{2}}}$. For this problem, the distance between two lines must be found. Since these two lines are parallel, pick a point alongside either one and apply the formula given. In this case, point $(0,1)$ on line $x+y=1$ will be used, so $(x_{1},y_{1})=(0,1)$. Take $5$ to the left side in the other equation for it to become $x+y-5=0$. Then we have that $A=1$, $B=1$ and $C=-5$ Substitute the known values into the formula and evaluate: Distance $=\dfrac{|(1)(0)+(1)(1)-5|}{\sqrt{1^{2}+1^{2}}}=...$ $...=\dfrac{|0+1-5|}{\sqrt{1+1}}=\dfrac{|-4|}{\sqrt{2}}=\dfrac{4}{\sqrt{2}}=...$ Rationalize the denominator: $...=\dfrac{4}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{4\sqrt{2}}{(\sqrt{2})^{2}}=\dfrac{4\sqrt{2}}{2}=2\sqrt{2}$
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