Question

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

Answer 1

We know the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is,

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

points are $A(2,4),B(2,6)andC(2+\sqrt{3},5)$

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

$BC=\sqrt{(2+\sqrt{3}-2{)}^2+(5-6{)}^2}=\sqrt{3+1}=\sqrt{4}=2$

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

AB = BC = CA
∴ the vertices are of an equilateral triangle