Question

Show that A(1, 2), (1, 6), C(1 + 2 ${\displaystyle \sqrt{3}}$, 4) are vertices of a equilateral triangle

Answer 1

Distance between two points = ${\displaystyle \sqrt{{(x_2- x_1)}^2+{(y_2-y_1)}^2}}$

By distance formula,

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

= ${\displaystyle \sqrt{0^2+4^2}}$

= ${\displaystyle \sqrt{4^2}}$

= 4    ......(i)

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

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

= ${\displaystyle \sqrt{12+4}}$

= ${\displaystyle \sqrt{16}}$

= 4    .....(ii)

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

= ${\displaystyle \sqrt{{\left(2\sqrt{3}\right)}^2+2^2}}$

= ${\displaystyle \sqrt{12+4}}$

= ${\displaystyle \sqrt{16}}$

= 4   ......(iii)

∴ AB = BC = AC   ......[From (i), (ii) and (iii)]

∴ ∆ABC is an equilateral triangle.

∴ Points A, B and C are the vertices of an equilateral triangle.