Question

Show that the points A(1, 2), B(1, 6), C(1 + 2`sqrt3`, 4) are vertices of an equilateral triangle.

Answer 1

The given points are A(1, 2), B(1, 6), C(1 + 2`sqrt3`, 4).

`"Distance between" = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`

By distance formula,

AB = `sqrt((1 - 1)^2 + (6 - 2)^2)`

∴ AB = `sqrt((0)^2 + (4)^2)`

∴ AB = `sqrt(0 + 16)`

∴ AB = `sqrt(16)`

∴ AB = 4        ...(1)

BC = `sqrt((1 + 2sqrt3 - 1)^2 + (4 - 6)^2)`

∴  BC = `sqrt((2sqrt3)^2 + (-2)^2)`

∴  BC = `sqrt(12 + 4)`

∴  BC = `sqrt(16)`

∴  BC = 4        ...(2)

AC = `sqrt((1 + 2sqrt3 - 1)^2 + (4 - 2)^2)`

∴  AC = `sqrt((2sqrt3)^2 + (2)^2)`

∴  AC = `sqrt(12 + 4)`

∴  AC = `sqrt(16)` 

∴  AC = 4        ...(3)

From (1), (2) and (3)

∴  AB = BC = AC = 4

Since, all the sides of equilateral triangle are congruent.

∴ ΔABC is an equilateral triangle.

The points A, B and C are the vertices of an equilateral triangle.