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Question
Show that the points (-3, -3),(3,3) and C (-3 `sqrt(3) , 3 sqrt(3))` are the vertices of an equilateral triangle.
Solution
Let the given points be(-3, -3),(3,3) and C (-3 `sqrt(3) , 3 sqrt(3))` Now
`AB = sqrt((-3-3)^2 +(-3-3)^2) = sqrt((-6)^2 +(-6)^2)`
`= sqrt(36+36) = sqrt(72) = 6sqrt(2)`
`BC = sqrt((3+3sqrt(3))^2 +(3-3sqrt(3))^2)`
`= sqrt(9+27+18sqrt(3) +9+27-18sqrt(3)) = sqrt(72)=6 sqrt(2)`
`AC = sqrt((-3+3sqrt(3))^2 +(-3-3sqrt(3))^2 )= sqrt((3-3sqrt(3))^2 +(3+3sqrt(3))^2)`
`= sqrt(9+27-18sqrt(3)+9+27+18sqrt(3)`
`= sqrt(72)=6sqrt(2)`
Hence, the given points are the vertices of an equilateral triangle.
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