## Calculus 8th Edition

$x=4t,y=3t,z=-t$ and $\frac{x}{4}=\frac{y}{3}=\frac{z}{-1}$
The direction vector for a line through the origin $(0,0,0)$ and the point $(4,3,-1)$ is $\lt 4,3,-1 \gt$ . Parametric equations defined by: $x=x_0+at$, $y=y_0=bt$ and $z=z_0+ct$ Thus, the parametric equations are: $x=4t,y=3t,z=-t$ 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: $\frac{x}{4}=\frac{y}{3}=\frac{z}{-1}$