Question

Show that the points $A(1,2),B(1,6),C(1+2\sqrt{3},4)$ are vertices of an equilateral triangle.

Answer 1

The given points are $A(1,2),B(1,6),C(1+2\sqrt{3},4)$

The distance between any two point is given by (Distance Formula)

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$AB=\sqrt{(1-1{)}^2+(2-6{)}^2}=\sqrt{0+(-4{)}^2}=\sqrt{16}=4$

$BC=\sqrt{(1-1-2\sqrt{3}{)}^2+(6-4{)}^2}=\sqrt{12+4}=\sqrt{16}=4$

$AC=\sqrt{(1-1-2\sqrt{3}{)}^2+(2-4{)}^2}=\sqrt{12+4}=\sqrt{16}=4$

$\therefore AB=BC=AC=4$
Since, all the sides of the triangle formed by the points A, B and C are equal so, the triangle formed is equilateral triangle.