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