Area of equilateral ∆ is √3/4 a^2......(1)
Also area of ∆ is 1/2{|X1-X2|+
|X2-X3|+|X3-X1|}........(2)
If coordinates are rational then it's area will be rational but from (1) it is irrational.
so ∆ is not possible

Proof by contradiction:

Assumption: Suppose there exists an equilateral triangle with vertices having rational coordinates.

Properties of an equilateral triangle:
1. All three sides of an equilateral triangle are congruent.
2. All three angles of an equilateral triangle are congruent and each measures 60 degrees.

Proof:
Let's assume that the vertices of the equilateral triangle are A(x1, y1), B(x2, y2), and C(x3, y3), where x1, x2, x3, y1, y2, y3 are rational numbers.

1. Side lengths:
The distance between two points (x1, y1) and (x2, y2) is given by the distance formula:
d = √((x2 - x1)^2 + (y2 - y1)^2)

Since all sides of an equilateral triangle are congruent, the distances AB, BC, and AC should be equal. Let's assume that AB = BC = AC = d.

Substituting the coordinates, we have:
AB = √((x2 - x1)^2 + (y2 - y1)^2)
BC = √((x3 - x2)^2 + (y3 - y2)^2)
AC = √((x3 - x1)^2 + (y3 - y1)^2)

Since d is a rational number, the square of d must also be a rational number. Therefore, the square of each side length must be a rational number.

AB^2 = (x2 - x1)^2 + (y2 - y1)^2
BC^2 = (x3 - x2)^2 + (y3 - y2)^2
AC^2 = (x3 - x1)^2 + (y3 - y1)^2

The sum of any two rational numbers is also a rational number. Therefore, AB^2 + BC^2 and AB^2 + AC^2 should be rational numbers.

AB^2 + BC^2 = (x2 - x1)^2 + (y2 - y1)^2 + (x3 - x2)^2 + (y3 - y2)^2 = 2(x2 - x1)^2 + 2(y2 - y1)^2
AB^2 + AC^2 = (x2 - x1)^2 + (y2 - y1)^2 + (x3 - x1)^2 + (y3 - y1)^2 = 2(x2 - x1)^2 + 2(y2 - y1)^2

Since the sum of two rational numbers is rational, 2(x2 - x1)^2 + 2(y2 - y1)^2 should also be rational. Therefore, AB^2 + BC^2 = AB^2 + AC^2.

However, this implies that BC^2 = AC^2, which means the lengths of sides BC and AC are equal. This contradicts the assumption that all three sides are congruent in an equilateral triangle. Hence, an equilateral triangle cannot have all vertices with rational coordinates.

Therefore, the assumption that an equilateral triangle with rational coordinates exists is false. QED