## Calculus (3rd Edition)

$$c(t)=(x(t),y(t))=(2+4t,5-t).$$
Since the line is perpendicular to the line $y=4x-3$, then its slope is $m=-\frac{1}{4}$. Then we have $\frac{s}{r}=-\frac{1}{4}$, so take $r=4$ and $s=-1$. Then, the parametric equations are $$x=a+rt=2+4t, \quad y= b+st=5-t.$$ That is, $$c(t)=(x(t),y(t))=(2+4t,5-t).$$