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Show that the quadrilateral ABCD formed by joining the points A($-2,-1$), B($1, -1$), C($1, 2$) and D($-2, 2$) in order is a square.
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$AB = \sqrt{3^2 + 0^2} = 3$; $BC = \sqrt{0^2 + 3^2} = 3$
$CD = \sqrt{3^2 + 0^2} = 3$; $AD = \sqrt{0^2 + 3^2} = 3$ (1/2 $\times 4 = 2$ Marks)
$AC = \sqrt{3^2 + 3^2} = \sqrt{18}$ or $3\sqrt{2}$
$BD = \sqrt{3^2 + 3^2} = \sqrt{18}$ or $3\sqrt{2}$ (1/2 $\times 2 = 1$ Mark)
All sides are equal and diagonals are equal, hence ABCD is a square. (1/2 Mark)
$CD = \sqrt{3^2 + 0^2} = 3$; $AD = \sqrt{0^2 + 3^2} = 3$ (1/2 $\times 4 = 2$ Marks)
$AC = \sqrt{3^2 + 3^2} = \sqrt{18}$ or $3\sqrt{2}$
$BD = \sqrt{3^2 + 3^2} = \sqrt{18}$ or $3\sqrt{2}$ (1/2 $\times 2 = 1$ Mark)
All sides are equal and diagonals are equal, hence ABCD is a square. (1/2 Mark)