Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Section 12.5 - Equations of Lines and Planes - 12.5 Exercises - Page 832: 60

Answer

$\frac{x}{2}=\frac{y}{-1}=\frac{z+5}{5}$ or: $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z}{5}$

Work Step by Step

Given: $z=2x-y-5$ and $z=4x+3y-5$ Here, $(x_0,y_0,z_0)=(0,0,-5)$ and $\lt a,b,c\gt=\lt 4,-2,10\gt$ The symmetric equations are defined by: $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ Hence, the symmetric equations are: $\frac{x}{2}=\frac{y}{-1}=\frac{z+5}{5}$ or: $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z}{5}$
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