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Question
Show that the following points are the vertices of a square:
A (0,-2), B(3,1), C(0,4) and D(-3,1)
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Solution
The given points are A (0,-2), B(3,1), C(0,4) and D(-3,1)
`AB = sqrt ((3-0)^2 +(1+2)^2) = sqrt((3)^2+(3)^2) = sqrt(9+9) = sqrt(18) = 3sqrt(2) units`
`BC = sqrt ((0-3)^2 +(4-1)^2) = sqrt((-3)^2 +(3)^2) = sqrt(9+9) = sqrt(18) = 3 sqrt(2) units`
`CD = sqrt((-3-0)^2 + (1-4)^2) = sqrt((-3)^2 +(-3)^2 ) = sqrt(9+9) = sqrt(18) = 3 sqrt(2) units`
`DA = sqrt((-3-0)^2 +(1+2)^2) = sqrt((-3)^2 +(3)^2) = sqrt(9+9) = sqrt(18) = 3 sqrt(2) units`
Therefore, `AB = BC = CD = DA = 3 sqrt(2) units`
Also ,
`AC= sqrt((0-0)^2 + (4+2)^2) = sqrt((0)^2 +(6)^2 ) = sqrt(36) = 6 units`
`BD = sqrt((-3-3)^2 +(1-1)^2) = sqrt((-6)^2 +(0)^2) = sqrt(36) =6 units`
Thus, diagonal AC = diagonal BD
Therefore, the given points from a square.
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