## Precalculus (6th Edition) Blitzer

$x=6t+3$ ; $y=13t-1$
For a line, we have: $(3,-1); (9,12)$ $x_1=3, x_2=9,y_1=-1, y_2=12$ Now, in parametric from: $x=(x_2-x_1) t+x_1=6t+3$ and $y=(y_2-y_1)t_1+y_1 =13t-1$ So, $x=6t+3$ ; $y=13t-1$