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Question
Prove that the points (2a, 4a), (2a, 6a) and `(2a + sqrt3a, 5a)` are the vertices of an equilateral triangle.
Solution
The distance d between two points `(x_1, y_1)` and `(x_2, y_2)` is given by the formula
`d = sqrt((x_1 + x_2)^2 + (y_1- y_2)^2)`
In an equilateral triangle all the sides have equal length.
Here the tree points are A(2a, 4a), B(2a, 6a) and `C(2a + asqrt3, 5a)`
Let us now find out the lengths of all the three sides of the given triangle.
`AB = sqrt((2a - 2a)^2 + (4a - 6a)^2)`
`= sqrt((0)^2 + (-2a)^2)`
`= sqrt((0)^2 + (-2a)^2)`
`= sqrt(0 + 4a^2)`
AB = 2a
`BC = sqrt((2a - 2a - asqrt3)^2 + (6a - 5a))`
`= sqrt((-asqrt3)^2 + (a)^2)`
`= sqrt(3a^2 + a^2)`
`= sqrt(4a^2)`
AC = 2a
Since all the three sides have equal lengths the triangle has to be an equilateral triangle.
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