If the vertices of a triangle have rational coordinates, then prove that the triangle cannot be equilateral.

Explanation

Let the vertices of triangle ABC are `A(x_1,y_1) ,B (x_2,y_2)` and `C(x_3,y_3)`.
Since vertices have rational coordinates, consider `x_i,y_iepsilonQ` for `i=1,2,3`.
Area of triangle, `Delta=(1)/(2) [(x_1y_2-x_2y_1)+(x_2y_3-x_3y_2)+(x_3y_1-x_1y_3)]` (1)
Clerly, this value will be rational. Also, area of equilateral triangle, `Delta=sqrt(3)/(4)a^2`, where a is side length.
`a=AB=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
`therefore" "Delta=sqrt(3)/(4)[(x_2-x_1)^2+(y_2-y_1)^2]` (2)
Clearly,this value is irrational. Thus, we have contradiction. So, triangle cannot be equilateral.