#### Answer

$$c(t)= (2+t,1-2t)$$
(Other parametrizations are possible.)

#### Work Step by Step

Since the equation of the line is $y=5-2x$, then its slope is $m=-2$. Then we have $\frac{s}{r}=-2$, so take (for example) $r=1$ and $s=-2$. The line passes through one (sample) point $(0,5)$. Thus, the parametric equations are
$$x=a+rt=t, \quad y= b+st=5-2t.$$
That is,
$$c(t)=(x(t),y(t))=(t,5-2t).$$
Now, we define new parametrization $$c_0(s)=(s+h,5-2s-2h),$$
Since we need $c_0(0)=(2,1)=(h,5-2h)$, then $h=2$.
Hence the required parametrization is $$c_0(s)= (2+s,1-2s)$$
We rewrite the result in the form of $c(t)$:
$$c(t)= (2+t,1-2t)$$
(Other parametrizations are possible.)