NTSE Stage 2 SAT 2018 Exam Question Paper

Let the vertex be $\left({\mathit{x}}_{\mathit{r}},{\mathit{y}}_{\mathit{r}}\right),\mathit{r}=1,2,3$, where both ${\mathit{x}}_{\mathit{r}}$ and ${\mathit{y}}_{\mathit{r}}$ are integers.
Hence, area of triangle
$=\frac{1}{2}\mathit{\Sigma }{\mathit{x}}_1\left({\mathit{y}}_2-{\mathit{y}}_3\right)=$rational number .....(1)
Let a be the side of equilateral triangle, then
${\mathit{a}}^2={\left({\mathit{x}}_1-{\mathit{x}}_2\right)}^2+{\left({\mathit{y}}_1-{\mathit{y}}_2\right)}^2=$a positive integer.
But, the area of an equilateral triangle $=\frac{\sqrt{3}}{4}{\mathit{a}}^2$
∴ Area $=\left(\frac{\sqrt{3}}{4}\right){\mathit{a}}^2$,which is irrational, since ${\mathit{a}}^2$ is a positive integer.
Thus the two statements (1) and (2) for area are contradictory.
Hence, such type of a triangle is not possible.