Show that the points A(1,2),B(1,6), `C(1 + 2sqrt(3),4) ` are the
vertices of an equilateral triangle .

Explanation

`A(1,2),B(1,6),C(1 + 2sqrt(3),4)`
By distance formula ,
` AB = sqrt((1 - 1)^(2) + (6-2)^(2))`
` therefore AB = sqrt(0^(2) + 4^(2))`
` therefore AB= sqrt(4^(2))`
` therefore AB = 4 `
`BC = sqrt((1 + 2sqrt(3)-1)^(2) + (4 -6)^(2))`
` therefore BC= sqrt((2sqrt(3)^(2) + )(-2)^(2))`
` therefore BC = sqrt(12 + 4)`
` therefore BC = sqrt(16) `
`therefore BC = 4`
`therefore AC = sqrt((1 + 2sqrt(3)-1)^(2) + (4-2)^(2))`
` therefore AC = sqrt((2sqrt(3))^(2) + (2)^(2))`
` therefore AC = sqrt(12 + 4)`
` therefore AC = sqrt(16)`
` therefore AC = 4 `
` therefore AB = BC = AC `
` therefore Delta ABC ` is an equilateral triangle .i.e. the points A , B and C ar the
vertices of an equilateral triangle .