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Question
Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.
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Solution
We use the distance formula = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)` to find the length of each side.
AB = `sqrt((1 - 5)^2 + (5 - 6)^2)`
= `sqrt((-4)^2 + (-1)^2)`
= `sqrt(16 +1)`
= `sqrt(17)`
BC = `sqrt((2 - 1)^2 + (1 - 5)^2)`
= `sqrt((1)^2 + (-4)^2)`
= `sqrt(1+16)`
= `sqrt(17)`
CD = `sqrt((6 - 2)^2 + (2 - 1)^2)`
= `sqrt((4)^2 + (1)^2)`
= `sqrt(16 + 1)`
= `sqrt(17)`
DA = `sqrt((5 - 6)^2 + (6 - 2)^2)`
= `sqrt((-1)^2 + (4)^2)`
= `sqrt(1+16)`
= `sqrt(17)`
To prove it is a square, the diagonals must also be equal in length.
AC = `sqrt((2 - 5)^2 + (1 - 6)^2)`
= `sqrt((-3)^2 + (-5)^2)`
= `sqrt(9+25)`
= `sqrt(34)`
BD = `sqrt((6 - 1)^2 + (2 - 5)^2)`
= `sqrt((5)^2 + (-3)^2)`
= `sqrt(25 + 9)`
= `sqrt(34)`
The diagonals have the same length, `sqrt(34)` units.
Since, AB = BC = CD = DA and AC = BD,
A, B, C and D are the vertices of a square.
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