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  • If the vertices of a triangle have integral coordinates, then the triangle:  Option: 1 cannot be equilateral

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

 

Option: 1

cannot be equilateral


Option: 2

may or may not be equilateral 


Option: 3

always equilateral


Option: 4

is not possible 


Answers (1)

best_answer

Let \mathrm{A\equiv \left ( x_{1},y_{1} \right ), B\equiv \left ( x_{2}, y_{2} \right ),C\equiv \left ( x_{3} ,y_{3}\right )} be a triangle and \mathrm{x_{1},x_{2}, x_{3}, y_{1}, y_{2}, y_{3}} be integers. 
\mathrm{\therefore \mathrm{BC}^2=\left(\mathrm{x}_2-\mathrm{x}_3\right)^2+\left(\mathrm{y}_2-\mathrm{y}_3\right)^2} a positive integer. 
If the triangle is equilateral then \mathrm{\mathrm{AB}=\mathrm{BC}=\mathrm{CA}=\mathrm{a} \text { (say) and } \angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=60^{\circ}.}
\mathrm{\mathrm{\therefore } Area of the triangle \mathrm{=(1 / 2) \mathrm{bc}\ \sin \mathrm{A}=(1 / 2) \mathrm{a}^2 \sin\ 60^{\circ}=(\sqrt{3} / 2)=(\sqrt{3} / 4) \mathrm{a}^2} which is irrational, \mathrm{\because } \mathrm{a^{2} } is a positive integer.
Now, the area of the triangle in terms of the coordinates \mathrm{ =(1 / 2)\left[\mathrm{x}_1\left(\mathrm{y}_2-\mathrm{y}_3\right)+\mathrm{x}_2\left(\mathrm{y}_3-\mathrm{y}_1\right)+\mathrm{x}_3\left(\mathrm{y}_1-\mathrm{y}_2\right)\right], } Which is a rational number. This contradicts that the area is an irrational number, if the triangle is equilateral. 

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sudhir kumar

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