Question

Find the number of points on the straight line which points $(-4,11)$ to $(16,-1)$ whose coordinates are positive integer.

Answer 1

The slope of the line is,

$\frac{-1-11}{16-(-4)}=\frac{-12}{20}=\frac{-3}{5}$

Thus, the eqn of the line is,

$y-11=\frac{-3}{5}(x-(-4\left)\right)=\frac{-3}{5}x$

$\frac{-12}{5}=\frac{-1}{5}(3x+12)$

$y$ is am integer precisely when $3x+12$ is a multiply of $5$, which is when $(x+4)$ is a multiple of $5$

These values of $x$ occur $5$ units apast, between $-4$ and $6$ they arn't $-4,1,6,11$ and $16$ and corresponding $y-$ coordinates are $11,8,5,2$ and $-1$.

They values of $x$ between $-4$ and $6$ that give integral values of $y$ must be $-4,1,6,11$ and $16$. The corresponding $y-$ coordinates are $11,8,5,2$ and $-1$.

The line segment constants exactly $3$ points whose coordinates are path positive integers.