I'm Luigi!! Number one!!

Consider the triangle in the cartesian plane with $A(0;0)$ , $B(8;0)$ and $C(0;6)$ .

The equation of the the line which pass throught B and C is $3x+4y-24=0$

Call $P(x_{0};y_{0})$ a point inside the triangle with $x_{0}$ and $y_{0}$ integer so that the distance from AB and AC satisfy the problem.

The distance between P and the line BC is $d=\frac {|3x_{0}+4y_{0}-24|}{\sqrt{3^2+4^2}}$ .

At this point we have to find all the points $P(x_{0};y_{0})$ so that also $d=\frac {|3x_{0}+4y_{0}-24|}{5}$ is an integer.

There are only $\boxed{3}$ solutions (after having tried all the few points inside the triangle) which are $X(5;1)$ , $Y(2;2)$ and $Z(1;4)$ .

the triangle will be a right angle triangle with the hypotenuse 10 . let , the equation of the hypotenuse be 8x+6y=48 or 6x+8y=48 . sorry ! it seems to be uncomfortable to explain further . the only thing i can say that is i have used co-ordinate geometry formulae .