Answer
$x=2+2t,y=1+\frac{1}{2}t,z=-3-4t$
and
$\frac{x-2}{2}=2y-2=\frac{z+3}{-4}$
Work Step by Step
The direction vector for a line through the $(0,1/2,1)$ and the point $(2,1,-3)$ is $\lt 2,1/2,-4 \gt$ .
Parametric equations defined by:
$x=x_0+at$, $y=y_0=bt$ and $z=z_0+ct$
Thus, the parametric equations are:
$x=2+2t,y=1+\frac{1}{2}t,z=-3-4t$
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:
$\frac{x-2}{2}=\frac{y-1}{1/2}=\frac{z+3}{-4}$
$\frac{x-2}{2}=2y-2=\frac{z+3}{-4}$
