The equation of the line passing through (1,11) and (4,8) can be expressed in the form
x/a + y/b = 1.
Find a.
Let $\gamma = \frac{1}{a}$ and $\theta = \frac{1}{b}.$
We have $\gamma x + \theta y = 1$
We have $\gamma \cdot 1 + \theta \cdot 11 = 1$
We also have $\gamma \cdot 4 + \theta \cdot 8 = 1$
Multiplying first equation by $4,$ we have $4 \gamma + 44 \theta = 4$
Subtracting, we have $36 \theta = 3$
$\theta = \frac{1}{12}$
$\gamma = \frac{11}{12}$
Thus, $a = \frac{1}{\gamma} = \frac{1}{\frac{11}{12}} = \boxed{\frac{12}{11}}$