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Question
Prove that the points A(1, 7), B (4, 2), C(−1, −1) D (−4, 4) are the vertices of a square.
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Solution
The distance d between two points `(x_1, y_1)` and `(x_2, y_2)` is given by the formula
`d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)`
In a square, all the sides are equal in length. Also, the diagonals are equal in length in a square.
Here the four points are A(1, 7), B(4, 2), C(−1, −1) and D(−4, 4).
First, let us check if all the four sides are equal.
`AB = sqrt((1- 4)^2 + (7 - 2)^2)`
`= sqrt((-3)^2 + (5)^2)`
`= sqrt(9 + 25)`
`BC = sqrt34`
`CD = sqrt((-1 + 4)^2 + (-1-4)^2)`
`= sqrt((3)^2 + (-5)^2 )`
`= sqrt(9 + 25)`
`CD = sqrt(34)`
`AD = sqrt((1 +4)^2 + (7 - 4)^2)`
`= sqrt((5)^2 + (3)^2)`
`= sqrt(25 + 9)`
`AD = sqrt34`
Since all the sides of the quadrilateral are the same it is a rhombus.
For the rhombus to be a square the diagonals also have to be equal to each other.
`AC = sqrt((1 + 1)^2 + (7 + 1)^2)`
`=sqrt((2)^2 + (8)^2)`
`=sqrt(4 + 64)`
`AC = sqrt(68)`
`BD = sqrt((4 + 4)^2 + (2 + 4)^2)`
`= sqrt((8)^2 + (-2)^2)`
`= sqrt(64 + 4)`
`BD = sqrt(68)`
Since the diagonals of the rhombus are also equal to each other the rhombus is a square.
Hence the quadrilateral formed by the given points is a square.
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