Area of triangle with vertex $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by :
$$\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1 \end{vmatrix}$$
In this determinant if we take all the coordinates as rational numbers, we will never get an irrational number as an answer.
Does that mean a triangle's area can't ever be irrational if its coordinates are rational? (because I don't think so)
Also,is there anything even remotely similar in 3d to the shoelace formula?
Best Answer
Yes indeed, a direct consequence of the shoelace formula is that a triangle (convex polygon) having all its vertices with rational coordinates has a rational area. It is a nice technique for showing that there is no equilateral triangle in $\mathbb{Z}\times\mathbb{Z}$, too.