Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.1 Parametric Equations - Exercises - Page 603: 32

$$x= \frac{1}{3}+t, \quad y= \frac{1}{6}-t.$$

First, we find the slope $m=\frac{\frac{10}{6}-\frac{1}{6}}{-\frac{7}{6}-\frac{2}{6}}=\frac{9}{-9}=-1$. Now, we have $m=\frac{s}{r}=-1$ so take $r=1$ and $s=-1$. Then, the parametric equations are $$x=a+rt=\frac{1}{3}+t, \quad y= b+st=\frac{1}{6}-t.$$ That is, $$c(t)=(x(t),y(t))=(\frac{1}{3}+t,\frac{1}{6}-t).$$