Show that the points A(a,a),B(-a,-a) and C(-a√3,a√3) form an equilateral triangle

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Solution

To show that it is an equilateral triangle we will show that the 3 sides AB, AC and BC are equal.
We will use the distance formula to calculate the legths of the sides.
if (x1,y1) and (x2,y2) are 2 points then the distance d between them is given by the formula
d= $\sqrt{(} (x 1 - x 2)^{2} + (y 1 - y 2)^{2})$ . Appllyinhg this formula we get:
AB = $\sqrt{(} (a - (- a))^{2} + (a - (- a)^{2})$
= $\sqrt{(} (a + a)^{2} + (a + a)^{2})$
= $\sqrt{(} (2 a)^{2} + (2 a)^{2})$
= $\sqrt{(} 4 a^{2} + 4 a^{2})$
= $\sqrt{(} 8 a^{2})$
= $2 a \sqrt{2}$
AC = $\sqrt{(} (a - (- a \sqrt{3}))^{2} + (a - a \sqrt{3})^{2})$
= $\sqrt{(} (a + a \sqrt{3})^{2} + (a - a \sqrt{3})^{2})$
= $\sqrt{(} a^{2} + 2 a \sqrt{3} + 3 a^{2} + a^{2} - 2 a \sqrt{3} + 3 a^{2})$
= $\sqrt{(} 8 a^{2})$
= $2 a \sqrt{2}$
BC = $\sqrt{(} (- a - (- a \sqrt{3}))^{2} + (- a - a \sqrt{3})^{2})$
= $\sqrt{(} (- a + a \sqrt{3})^{2} + (- a - a \sqrt{3})^{2})$
= $\sqrt{(} a^{2} - 2 a \sqrt{3} + 3 a^{2} + a^{2} + 2 a \sqrt{3} + 3 a^{2})$
= $\sqrt{(} 8 a^{2})$
= $2 a \sqrt{2}$
It is seen from above that AB=AC=BC and therefore ABC is an equilateral triangle.
Proved

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