### Instruction: The points $(1, 7), (13, 16)$ and $(5, k)$, where $k$ is an integer, are vertices of a triangle. What is the sum of the values of $k$ for which the area of the triangle is a minimum?### Response: To find the minimum area of a triangle with two fixed vertices, I want to place the third vertex on the line segment joining the two fixed vertices, since that will make the height of the triangle zero.
So I need to find the equation of the line that passes through $(1, 7)$ and $(13, 16)$.
Using the slope formula, I get that the slope of this line is $\frac{16 - 7}{13 - 1} = \frac{9}{12} = \frac{3}{4}$.
Using the point-slope form, I get that the equation of the line is $y - 7 = \frac{3}{4}(x - 1)$.
Simplifying, I get $y = \frac{3}{4}x + \frac{25}{4}$.
Now I want to find the value of $k$ that makes $(5, k)$ lie on this line.
Substituting $x = 5$ into the equation, I get $k = \frac{3}{4}(5) + \frac{25}{4} = \frac{40}{4} = 10$.
So the only value of $k$ that minimizes the area of the triangle is $10$.