Show that the points A(1, 2), B(1, 6), C(1 + 2 3 , 4) are vertices of an equilateral triangle.
The given points are A(1, 2), B(1, 6), C(1 + 2√3, 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.
Chapter 5. Co-ordinate Geometry – Practice Set 5.1 (Page 107)