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' repeatedly will give other points on the line that have coordinates which are both integers. These points are $(-9,-2),(-6,0),(-3,2),(0,4),(3,6),(6,8)$. Therefore, there are 6 such points.