### 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 minimize the area of the triangle, I want to make it as flat as possible, meaning the slope of the line through any two vertices should be the same.
I can use the slope formula to find the slope of the line through $(1, 7)$ and $(13, 16)$: $m = \frac{16 - 7}{13 - 1} = \frac{9}{12} = \frac{3}{4}$.
Now I want to find the values of $k$ that make the slope of the line through $(1, 7)$ and $(5, k)$ also equal to $\frac{3}{4}$.
I can use the slope formula again: $\frac{3}{4} = \frac{k - 7}{5 - 1}$.
Cross-multiplying, I get $3(5 - 1) = 4(k - 7)$.
Simplifying, I get $12 = 4k - 28$.
Adding $28$ to both sides, I get $40 = 4k$.
Dividing both sides by $4$, I get $k = 10$.
So one value of $k$ that minimizes the area of the triangle is $10$.
But is it the only one?
I can use the same logic to find the slope of the line through $(13, 16)$ and $(5, k)$ and set it equal to $\frac{3}{4}$.
I get $\frac{3}{4} = \frac{16 - k}{13 - 5}$.
Cross-multiplying, I get $3(13 - 5) = 4(16 - k)$.
Simplifying, I get $24 = 64 - 4k$.
Subtracting $64$ from both sides, I get $-40 = -4k$.
Dividing both sides by $-4$, I get $k = 10$.
So the same value of $k$ works for both lines.
Therefore, $k = 10$ is the only value of $k$ that minimizes the area of the triangle.
The sum of the values of $k$ is just $10$.