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(B) Let the vertices of the triangle be $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ where all coordinates are integers.The area of a triangle with inte

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Equilateral

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(B) Let the vertices of the triangle be $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ where all coordinates are integers.The area of a triangle with integer coordinates is given by the formula $	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$,which is always a rational number (specifically,a multiple of $0.5$).If the triangle were equilateral with side length $a$,its area would be $\frac{\sqrt{3}}{4} a^2$.Since the coordinates are integers,the square of the side length $a^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2$ must be an integer.Thus,the area $\frac{\sqrt{3}}{4} a^2$ would be an irrational number because $\sqrt{3}$ is irrational and $a^2$ is an integer.Since a triangle cannot have an area that is both rational and irrational,a triangle with integer coordinates can never be equilateral.

If the vertices of a triangle have integer coordinates,then what type of triangle can it never be?

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