Important question. No menu of choices given so we'll just have to figure it out.
It's not possible in two dimensions for all three vertices of an equilateral triangle to be lattice points, i.e. points with integer coordinates.
Since A(-2,1) and B(4,1) have integer coordinates, our other point won't. There will be a √3 involved, the algebraic tell that there's a 30/60/90 triangle lurking somewhere in the problem.
Here the given side is parallel to the x axis which makes things easier.
|AB| = 4 - -2 = 6
The x coordinate of the third vertex will be the same as that of the midpoint of AB, let's call it M,
M = ( (-2 + 4) / 2, (1 + 1)/2 ) = (1, 1)
We're making some progress. Whatever our height h is our two candidates for the third vertex C are (1, 1 ± h).
Since |AB|=6, we get |AB|=|BC|=|AC|=6 because it's an equilateral triangle.
The altitude CM bisects the triangle into 2 30/60/90 triangles. Let's take one of them, AMC. Angle AMC is a right angle, so we have a right triangle with legs |AM|=3 and |CM|=h, and hypotenuse |AC|=6. The Pythagorean Theorem tells us:
3² + h² = 6²
9 + h² = 36
h² = 27
h = 3√3
Answer: C is (1, 1 + 3√3) or (1, 1 - 3√3)
