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Question
Show that the following points are the vertices of a square:
(i) A (3,2), B(0,5), C(-3,2) and D(0,-1)
Solution
The given points are A (3,2), B(0,5), C(-3,2) and D(0,-1).
`AB = sqrt((0-3)^2 +(5-2)^2 ) = sqrt((-3)^2 +(3)^2 ) = sqrt(9+9) = sqrt(18) = 3 sqrt(2) units `
`BC= sqrt((-3-0)^2 + (2-5)^2) = sqrt((-3)^2 +(3)^2) = sqrt(9+9) = sqrt(18) = 3 sqrt(2) ` units
`CD = sqrt((0+3)^2 + (-1-2)^2) = sqrt((3)^2 +(-3)^2) = sqrt(9+9) = sqrt(18) = 3 sqrt(2) units`
`DA = sqrt((0-3)^2 +(-1-2)^2) = sqrt((-3)^2+(-3)^2) = sqrt(9+9) =sqrt(18) = 3sqrt(2) units`
Therefore`AB =BC=CD=DA=3 sqrt(2) units`
Also, `AC = sqrt((-3-3)^2 +(2-2)^2) = sqrt((-6)^2 +(0)^2) = sqrt(36) = 6 units`
`BD = sqrt((0-0)^2 + (-1-5)^2) = sqrt((0)^2 +(-6)^2 )= sqrt(36) = 6 units`
Thus, diagonal AC = diagonal BD
Therefore, the given points from a square.
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