Question

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

Answer 1

Let A(x₁, y₁) B(x₂, y₂) and C(x₃, y₃) be vertices of a triangle (ABC) with integer values.
area of (Δ ABC) = $\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\}$
Since x_i, y_i; are integers
⇒ area of Δ(ABC) is rational ...(1)
Let us suppose Δ is an equilateral triangle
$$\begin{gathered} \mathrm{Then}\mathrm{\Delta }=\frac{\sqrt{3}}{4}{a}^2 \\ \mathrm{where}a=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2} \\ \mathrm{i}.\mathrm{e}.a^2={\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2 \\ \mathrm{i}.\mathrm{e}.a^2\mathrm{is}\mathrm{ration}. \\ \therefore \mathrm{\Delta }=\frac{\sqrt{3}}{4}{a}^2\mathrm{is}\mathrm{irrational} \end{gathered}$$
Which contradicts (1)
Hence Δ cannot be equilateral triangle.