Question

Prove that the points A(2, 4), b(2, 6) and (2 +${\displaystyle \sqrt{3}}$ ,5)  are the vertices of an equilateral triangle

Answer 1

The given points are A(2, 4), b(2, 6) and (2 +${\displaystyle \sqrt{3}}$ ,5) Now 

${\displaystyle AB=\sqrt{({(2-2)}^2+{(4-6)}^2)}=\sqrt{{\left(0\right)}^2+{(-2)}^2}}$

    ${\displaystyle =\sqrt{(0+4)=2}}$

${\displaystyle BC=\sqrt{{(2-2-\sqrt{3})}^2+{(6-5)}^2}=\sqrt{{(-\sqrt{3})}^2+{\left(1\right)}^2}}$

${\displaystyle =\sqrt{3+1}=2}$

${\displaystyle AC=\sqrt{{(2-2-\sqrt{3})}^2+{(4-5)}^2}=\sqrt{{(-\sqrt{3})}^2+{(-1)}^2}}$

${\displaystyle =\sqrt{3+1}=2}$

Hence, the points A(2, 4), b(2, 6) and (2 +${\displaystyle \sqrt{3}}$ ,5) are the vertices of an equilateral triangle