Including the endpoints, how many points on the line segment joining $(-9,-2)$ and $(6,8)$ have coordinates that are both integers?
(A) 2
(B) 7
(C) 16
(D) 11
(E) 6
Solution 1

The line segment with endpoints $(-9,-2)$ and $(6,8)$ has slope $\frac{8-(-2)}{6-(-9)}=\frac{10}{15}=\frac{2}{3}$.

This means that starting at $(-9,-2)$ and moving "up 2 and right 3" (corresponding to the rise and run of 2 and 3) repeatedly will give other points on the line that have coordinates which are both integers.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-067.jpg?height=526&width=534&top_left_y=160&top_left_x=1381)

These points are $(-9,-2),(-6,0),(-3,2),(0,4),(3,6),(6,8)$.

So far, this gives 6 points on the line with integer coordinates.

Are there any other such points?

If there were such a point between $(-9,-2)$ and $(6,8)$, its $y$-coordinate would have to be equal to one of $-1,1,3,5,7$, the other integer possibilities between -2 and 8 .

Consider the point on this line segment with $y$-coordinate 7 .

Since this point has $y$-coordinate halfway between 6 and 8 , then this point must be the midpoint of $(3,6)$ and $(6,8)$, which means that its $x$-coordinate is $\frac{1}{2}(3+6)=4.5$, which is not an integer.

In a similar way, the points on the line segment with $y$-coordinates $-1,1,3,5$ do not have integer $x$-coordinates.

Therefore, the 6 points listed before are the only points on this line segment with integer coordinates.

## Solution 2

The line segment with endpoints $(-9,-2)$ and $(6,8)$ has slope $\frac{8-(-2)}{6-(-9)}=\frac{10}{15}=\frac{2}{3}$.

Since the line passes through $(6,8)$, its equation can be written as $y-8=\frac{2}{3}(x-6)$ or $y=\frac{2}{3} x+4$. Suppose that a point $(x, y)$ lies along with line between $(-9,-2)$ and $(6,8)$ and has both $x$ and $y$ integers.

Since $y$ is an integer and $\frac{2}{3} x=y-4$, then $\frac{2}{3} x$ is an integer.

This means that $x$ must be a multiple of 3 .

Since $x$ is between -9 and 6 , inclusive, then the possible values for $x$ are $-9,-6,-3,0,3,6$.

This leads to the points listed in Solution 1, and justifies why there are no additional points.

Therefore, there are 6 such points.

ANSWER: (E)