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: 93

Answer

TRUE

Work Step by Step

Consider the first line $ax+by=c_{1}$: By subtacting $-ax$ from both sides, it can be rearranged to: $by=-ax+c_{1}$ Dividing $b$ from both sides gives the equation $y=-\frac{a}{b}x+\frac{c_{1}}{b}$. Therefore the gradient$=-\frac{a}{b}$ Consider the second line $bx-ay=c_{2}$: It can be rearranged to: $bx-c_{2}=ay$ Dividing both sides by $a$ gives the equation: $y=\frac{b}{a}x-\frac{c_{2}}{a}$ Therefore the gradient=$\frac{b}{a}$ Multiplying both gradients gives: $\frac{b}{a}\times-\frac{a}{b}=-1$ Therefore the lines must be perpendicular
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