If the vertices of a triangle have integral coordinates, then the triangle cannot be

Let A=(x1,y1),B=(x2,y2),C=(x3,y3)  be the vertices of a triangle and x1,x2,x3,y1,y2,y3  be integers.∴  BC2=(x2−x3)2+(y2−y3)2  a positive integer.If the triangle is equilateral, then AB=BC=CA=a(say)  and ∠A=∠B=∠C=600∴  Area of the triangle =(12)bcsinA         =(12)2sin600=(a22)(32)=(34)a2  which is irrational.Now, the area of the triangle in terms of the co-ordinates         =(12)[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]Which is a rational number.This contradicts that the area of the irrational number.∴The triangle cannot be equilateral