### Instruction: 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?### Response: I need to find the distance between each pair of vertices and then compare them to see which one is the longest.
I can use the distance formula, which is $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
For example, to find the distance between $(11,1)$ and $(2,3)$, I plug in $x_1 = 11$, $x_2 = 2$, $y_1 = 1$, and $y_2 = 3$ into the formula and get $d = \sqrt{(2-11)^2+(3-1)^2}$.
Simplifying, I get $d = \sqrt{81+4} = \sqrt{85}$.
I repeat this process for the other two pairs of vertices: $(11,1)$ and $(3,7)$, and $(2,3)$ and $(3,7)$.
For $(11,1)$ and $(3,7)$, I get $d = \sqrt{(3-11)^2+(7-1)^2} = \sqrt{64+36} = \sqrt{100} = 10$.
For $(2,3)$ and $(3,7)$, I get $d = \sqrt{(3-2)^2+(7-3)^2} = \sqrt{1+16} = \sqrt{17}$.
Comparing these three values, I see that the longest side is the one between $(11,1)$ and $(3,7)$, which has a length of $10$ units.