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: 93



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
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.