Finding the number of points on the straight line joining $(-4,11)$ and $(16,-1)$ Find the number of points on the straight line which joins $(-4,11)$ and $(16,-1)$ whose coordinates are positive integers.
a) $1$
b) $2$
c) $3$
d) $4$

The slope of this line is $$\frac{-1-11}{16-(-4)}=\frac{-12}{20}=-\frac35\;.$$ Thus, the equation of the line is $y-11=-\frac35(x-(-4))=-\frac35x-\frac{12}5=-\frac15(3x+12)$. Clearly $y$ is an integer precisely when $3x+12$ is a multiple of $5$, which is the case precisely when $x+4$ is a multiple of $5$. These values of $x$ occur $5$ units apart, so between $-4$ and $16$ they are $-4,1,6,11$, and $16$, and the corresponding $y$-coordinates are $11,8,5,2$, and $-1$.
You don’t actually need the equation of the line. The slope of $-\frac35$ means that every $1$-unit increase in $x$ causes a decrease in $y$ of $\frac35$. Thus, it takes a $5$-unit increase in $x$ to decrease $y$ by an integer (namely, $3$), and the values of $x$ between $-4$ and $16$ that give integral values of $y$ must be $-4,1,6,11$, and $16$. The corresponding $y$-coordinates are $11,8,5,2$, and $-1$.
By either approach we see that the line segment contains exactly $3$ points whose coordinates are both positive integers.
