First slide

Cartesian plane

Question

If two vertices of an equilateral triangle have integral coordinates then the third vertex will have

Moderate

A

coordinates which are irrational
B

atleast one coordinate which is irrational
C

coordinates which are rational
D

coordinates which are integers

Solution

Let the vertices of the equilateral triangle be (x₁, y₁), (x₂, y₂) and (x₃, y₃).
If none of x_i and y_i (i = 1, 2, 3) are irrational, then
area of $∆ = \frac{1}{2} |\begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix}| = r e t i o n a l$
But the area of an equilateral triangle
$= \frac{\sqrt{3}}{4} {(s i d e)}^{2} = i r r a t i o n a l$
Thus, the two statements are contradictory. Therefore, both the coordinates of the third vertex cannot be rational.

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