A triangle has vertices at coordinates $(11,1)$, $(2,3)$ and $(3,7)$.  What is the number of units in the length of the longest side of the triangle?
We must find the distance between each pair of points by using the distance formula.

The distance between $(11, 1)$ and $(2, 3)$ is $\sqrt{(11 - 2)^2 + (1 - 3)^2} = \sqrt{81 + 4} = \sqrt{85}$.

The distance between $(2, 3)$ and $(3, 7)$ is $\sqrt{(2 - 3)^2 + (3- 7)^2} = \sqrt{1 + 16} = \sqrt{17}$.

The distance between $(3, 7)$ and $(11, 1)$ is $\sqrt{(11 - 3)^2 + (1- 7)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$.

$10$ is larger than $\sqrt{85}$ and $\sqrt{17}$. Thus, the longest side of the triangle has length $\boxed{10}$.