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  4. The coordinates of the vertices of an equilateral triangle cannot be all

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Q. The coordinates of the vertices of an equilateral triangle cannot be all

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Solution:

Let $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ and $C\left(x_3, y_3\right)$ be the vertices of a $\triangle A B C$. If possible let $x_1, y_1, x_2, y_2, x_3, y_3$ be all rational.
Now, area of $\triangle A B C$
$=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$
$ =$ Rational ......(i)
Since, $\triangle A B C$ is equilateral.
$\therefore $ Area of $ \triangle A B C =\frac{\sqrt{3}}{4}(\text { side })^2$
$ =\frac{\sqrt{3}}{4}(A B)^2 $
$ =\frac{\sqrt{3}}{4}\left\{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2\right\} $
$ =$ Irrational ........(ii)
From Eqs. (i) and (ii), we get
Rational $=$ Irrational
which is contradiction.
Hence, $x_1, y_1, x_2, y_2, x_3, y_3$ cannot all be rational.