Question

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

• A. irrational
• B. rational
• C. real
• D. complex

Answer 1

Answer: (B)

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 $△ABC$. If possible let $x_1,y_1,x_2,y_2,x_3,y_3$ be all rational.
Now, area of $△ABC$
$=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$
$=$ Rational ......(i)
Since, $△ABC$ is equilateral.
$\therefore$ Area of $△ABC=\frac{\sqrt{3}}{4}(\text{ side }{)}^2$
$=\frac{\sqrt{3}}{4}(AB{)}^2$
$=\frac{\sqrt{3}}{4}\left\{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2\right\}$
$=$ 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.