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Question
Show that A(1, 2), (1, 6), C(1 + 2 `sqrt(3)`, 4) are vertices of a equilateral triangle
Solution
Distance between two points = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
By distance formula,
AB = `sqrt((1- 1)^2 + (6 - 2)^2`
= `sqrt(0^2 + 4^2)`
= `sqrt(4^2)`
= 4 ......(i)
BC = `sqrt((1 + 2sqrt(3) - 1)^2 + (4 - 6)^2`
= `sqrt((2sqrt(3))^2 + (-2)^2`
= `sqrt(12 + 4)`
= `sqrt(16)`
= 4 .....(ii)
AC = `sqrt((1 + 2sqrt(3) - 1)^2 + (4 -2)^2`
= `sqrt((2sqrt(3))^2 + 2^2`
= `sqrt(12 + 4)`
= `sqrt(16)`
= 4 ......(iii)
∴ AB = BC = AC ......[From (i), (ii) and (iii)]
∴ ∆ABC is an equilateral triangle.
∴ Points A, B and C are the vertices of an equilateral triangle.
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