As I mentioned in my comment above, the simple formula is:
$$\sum_{y=0}^{Y}\left(1+\left\lfloor \frac{yX}{Y}\right\rfloor\right)$$
If you use Pick's formula to count the points in $(0,0)-(0,Y)-(\left\lfloor X \right\rfloor,Y)$ with result $N$, then your result is $$N + \sum_{y=1}^Y \left(\left\lfloor\frac{yX}{Y}\right\rfloor - \left\lfloor \frac{y\left\lfloor X\right\rfloor}Y \right\rfloor\right)$$
The values in the sum are always $0$ and $1$.
Basically, this sum counts the number of values $y=1,..,Y$ such that there is a natural number $n$ such that $\frac{y\left\lfloor X\right\rfloor}{Y} < n \leq \frac{yX}{Y}$
I've had some time to consider the continued fraction approach, and it might improve the computation in some instances.
Expand $\frac X Y$ as a continued fraction, choosing one with an odd number of terms if $X$ is rational. Let $R_k=\frac{P_k}{Q_k}$ be the $k$th continue fraction. Then the number of positive lattice points such that $y\leq Y$ and $R_{2n}<\frac{x}{y}\leq R_{2n+2}$ can be computed as the number of pairs of positive integers $(u,v)$ such that $uQ_{2n} + vQ_{2n+1}\leq Y$ and $ua_{2n+2} \geq v$. Letting $v_1=\left\lfloor \frac{v-1}{a_{2n+2}}\right\rfloor$, this count is:
$$\sum_{v=1}^{\left\lfloor \frac {a_{2n+2}Y}{Q_{2n+2}}\right\rfloor}\sum_{u=v_1+1}^{\left\lfloor \frac{Y-vQ_{2n+1}}{Q_{2n}}\right\rfloor} 1 $$
The inner sum can be written as $\max\left(0,\left\lfloor \frac{Y-vQ_{2n+1}}{Q_{2n}}\right\rfloor - \left\lfloor \frac{v-1}{a_{2n+2}}\right\rfloor\right)$.
Note that when $a_{2n+2}=1$, this term simplifies to:
$$\left\lfloor \frac{Y-vQ_{2n+1}}{Q_{2n}}\right\rfloor - (v-1) = \left\lfloor\frac{Y-v(Q_{2n+1}+Q_{2n})}{Q_{2n}}\right\rfloor +1 = \left\lfloor \frac{Y-vQ_{2n+2}}{Q_{2n}}\right\rfloor + 1$$
But $R_0=P_0$ is an integer, so the number of non-negative lattice points $(x,y)$ with $y\leq Y$ and $x\leq R_0y$ is $1+Y+\sum_{j=1}^Y P_0j = 1+Y+P_0\frac{Y(Y+1)}2$.
So the total number of points is:
$$1+Y+P_0\frac{Y(Y+1)}2 + \sum_{n=1}^\infty \sum_{v=1}^{\left\lfloor \frac {a_{2n}Y}{Q_{2n}}\right\rfloor} \max(0,\left\lfloor \frac{Y-vQ_{2n-1}}{Q_{2n-2}}\right\rfloor - \left\lfloor \frac{v-1}{a_{2n}}\right\rfloor)$$
But note that the number of terms in the inner sum is zero if $\frac{Q_{2n}}{a_{2n}}>Y$, and $\frac{Q_{2n}}{a_{2n}}>Q_{2n-1}$so we can always write this as a finite sum, and, in particular, since $Q_k$ grows exponentially (at minimum, $Q_k$ is the $k$th Fibonacci number), the number of $n$ values is actually $O(\log Y)$.
The total pairs $n,v$ in this sum is about $\frac{2Y}{a_1}$, maximum. When $a_1=1$, this is more terms than the $Y+1$ terms for the "simple formula" above. However, notice the simple formula requires you to know $X$ to an unspecified level of precision, while this formula only requires integer arithmetic with values less than $Y$.
For example, if $X=\frac{Y}{N}$ for some integer $N>1$, then the continued fraction for $\frac{1}{N}$ we would choose is: $$0+\frac{1}{N-1+\frac{1}{1}}$$ to ensure the odd number of terms, and we would have $P_0=0$, $Q_0=1$, $Q_2=N$. This also gives us the case of $a_2=1$, so we can use the easier formula for the terms. Then this result would yield (remembering that $X=\frac{Y}{N}$:) $$1+Y+\sum_{v=1}^{\left\lfloor \frac{Y}{N}\right\rfloor} (1+Y-vN) = (1+Y)( 1+\left\lfloor X\right\rfloor) - \frac{N\left\lfloor X\right\rfloor(\left\lfloor X\right\rfloor +1)}2$$
The only test I've given this formula is $Y=10$ and $N=3$ (so $X=\frac{10}{3}$.) Both this formula and simple counting yields $26$, but I'd want some more testing for this in general.
An alternative formula came to me today.
$$1+Y+P_0\frac{Y(Y+1)}2 + \sum_{n=1}^\infty \sum_{v=1}^{\left\lfloor \frac Y {Q_{2n-1}}\right\rfloor}\left( \left\lfloor \frac{Y-vQ_{2n-1}}{Q_{2n-2}}\right\rfloor-\left\lfloor \frac{Y-vQ_{2n-1}}{Q_{2n}}\right\rfloor\right)$$
In particular, then, if $Q_2+Q_1>Y$, then this total is:
$$1+Y+P_0\frac{Y(Y+1)}2 + \sum_{v=1}^{\left\lfloor \frac Y {Q_1}\right\rfloor} Y-vQ_1 =$$
$$1+Y+P_0\frac{Y(Y+1)}2 + YY_0 - Q_1\frac{Y_0(Y_0+1)}2$$
Where $Y_0=\left\lfloor \frac Y {Q_1}\right\rfloor$.
For example, if $\frac X Y=\pi$, then $Q_1=7$, $Q_2=106$, so if $Y<113$, the total is:
$$1+Y+3\frac{Y(Y+1)}2 + YY_0 - 7\frac{Y_0(Y_0+1)}2$$
Where $Y_0=\left\lfloor \frac Y {7}\right\rfloor$.
Do a central reflection of your triangle in the midpoint of any one of its sides to get a new triangle that shares that side, but is otherwise disjoint from the original triangle; the two together form a parallelogram. One easily sees that the hypotheses about the triangle imply that parallelogram still has no interior lattice points, but its four vertices are lattice points.
To show intuitively that the parallelogram must have area $1$, pave the whole plane by translates of the parallelogram laid side-to-side in the obvious way. Now each lattice point is the bottom left corner (for some appropriate meaning of "bottom-left") of a unique parallelogram. Choosing a large convex region, so that the number of parallelograms crossing its boundary is small with respect to the number completely contained in it, the latter number is approximately equal to the number of lattice points in the region, which is approximately equal to the area of the region. Taking the limit as the size of the region goes to infinity, one sees that the area of individual parallelograms must be$~1$.
A somewhat more formal way to complete that argument is to roll up the plane to a torus: take the projection $\def\R{\Bbb R}\def\Z{\Bbb Z}\R^2\to\R^2/(n\Z\times m\Z)$ for some positive integers $n,m$. The codomain is a compact surface of area $nm$, which contains $nm$ points that are the image of a lattice points (element of $\Z\times\Z$). Choosing preimages for those points, and taking the corresponding set of parallelograms, one obtains a region that under the projection covers our compact surface exactly once; since the projection is locally area-preserving, that region has area $nm$, and each parallelogram has area$~1$. In fact one could just choose $n=m=1$.
Best Answer
Use Pick's theorem: The area of the figure is the number of internal lattice points plus half the boundary lattice points minus one.
$$ A = i + \frac{b}{2} - 1 $$
and also the result of another question here to count points on a line segment. Use this to count the points on the three boundary segments (being sure to count the points on the vertices only once) to get $b$.