Final answer:
An equilateral triangle cannot have integer coordinates for its vertices because the distance from a given vertex to the opposite side forms an altitude that involves the square root of 3, an irrational number, making it impossible for all coordinates to be integers.
Step-by-step explanation:
We are to explain why it is not possible to construct an equilateral triangle with three vertices having integer coordinates. Let's start by considering the definition of an equilateral triangle - it is a triangle in which all three sides are of equal length and all three angles are equal, each measuring 60 degrees.
Suppose we have two vertices of an equilateral triangle at integer coordinates, say (0,0) and (a,0), where 'a' is an integer. The distance between these points, which is the side length of our equilateral triangle, is 'a' units. The third vertex must also be at a distance of 'a' units from both of the other vertices. Using the Pythagorean theorem we can set up an equation for the coordinates of the third vertex, say (x,y), that would satisfy the conditions of integer coordinates and the distance 'a':
x2 + y2 = a2 and (x-a)2 + y2 = a2.
After solving these equations, we would expect the coordinates to be integers. However, due to the nature of square roots and the fact that the triangle's altitude forms a 30-60-90 triangle, the vertical coordinate (y) inevitably involves the square root of 3. Since the square root of 3 is an irrational number, 'y' cannot be an integer. Therefore, it is not possible to have an equilateral triangle with vertices at integer coordinates.
An example would be if 'a' were 2, implying side lengths of 2. The expected coordinates for the third vertex would be (1, sqrt(3)), with sqrt(3) being an irrational number and not an integer, contradicting the initial condition.