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)
\[BC = \sqrt{((1 + 2sqrt3 - 1)^2 + (4 - 6)^2)}\]
\[∴ BC = \sqrt{((2\sqrt3)^2 + (-2)^2)}\]
\[∴ BC = \sqrt{(12 + 4)}\]
\[∴ BC = \sqrt{(16)}\]
∴ BC = 4 ...(2)
\[AC = \sqrt{((1 + 2\sqrt3 - 1)^2 + (4 - 2)^2)}\]
\[∴ AC = \sqrt{((2sqrt3)^2 + (2)^2)}\]
\[∴ AC = \sqrt{(12 + 4)}\]
\[∴ AC = \sqrt{(16) }\]
∴ 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)