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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 the triangle is give

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Never 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 the triangle 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)|$.Since all coordinates are integers,the area must be a rational number (specifically,a multiple of $0.5$).If the triangle were equilateral with side length $a$,then $a^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2$. Since the coordinates are integers,$a^2$ must be a positive integer.The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} a^2$.Since $a^2$ is an integer,$\frac{\sqrt{3}}{4} a^2$ is an irrational number.This creates a contradiction because the area cannot be both rational and irrational. Therefore,a triangle with integral coordinates can never be equilateral.

If the vertices of a triangle have integral coordinates,then the triangle is

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