First slide

Cartesian plane

Question

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

Easy

A

integral coordinates
B

coordinates which are rational
C

at least one coordinate irrational
D

coordinates which are irrational

Solution

Let $A (x_{1}, y_{1}), B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ be the vertices of an equilateral triangle ABC such that $x_{1}, x_{2}$ and $y_{1}, y_{2}$ are integers. If we assume that none of the coordinates of the vertex C are irrational, then we find that
$Δ = Area of Δ A B C$
$\Rightarrow Δ = \frac{1}{2} |\begin{array}{lll} x_{1} y_{1} 1 \\ x_{2} y_{2} 1 \\ x_{3} y_{3} 1 \end{array}| = A$ rational number
But, $Δ = \frac{\sqrt{3}}{4} (Side)^{2} = \frac{\sqrt{3}}{4} \times A$ rational number
$\Rightarrow Δ =$ an irrational number.
Thus, we arrive at a contradiction. Therefore, our supposition is wrong.
Hence, at least one coordinate of C is irrational

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