Will you count?

In General
2 (no. of integral point)=(n-1)(n-2) here, n=6
no. of integral point=10

Triangle's area is $18$ due to is a rectangle triangle of cathets $6$

By Pick's Theorem

$\boxed{\text{Area}=I+\frac{B}{2}-1}$

$I=$ number of points in the interior of the polygon

$B=$ number of points on the boundary of the polygon

Counting $B$

On the vertical are (0,0), (0,1), (0,2), (0,3), (0,4), (0,5), (0,6)

On the horizontal are (0,0), (1,0), (2,0), (3,0), (4,0), (5,0), (6,0)

On the diagonal are (0,6), (1,5), (2,4), (3,3), (2,4), (1,5), (0,6),

$B=18$

Now, substituing

$18=I+\frac{18}{2}-1 \therefore \boxed{I=10}$

In the the interior of any triangle with coordinates as (0,0) , (n,0) and (0,n). The number of integer coordinates would be

(1,1) (1,2) ........... (1, n-2) (2,1) ...............(2,n-3) And so on upto (n-2,1)
Thus the total no. of integer coordinates would be 1 + 2 + 3.........n-2. Which is equal to (n-2)(n-1)/2 Putting in n as 6 here we get the desired answer which is 10.