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

Using distance formula,

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

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

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

Since $AB=BC=CA$

$\therefore ABC$ is an equilateral triangle