## Calculus (3rd Edition)

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