Show that following points are vertices of a square:
(0,1) , (1,4) , (4,3) , (3,0).

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Solution

$Let A (0, 1); B (1, 4); C (4, 3); D (3, 0) are the given points . Now, we know that distance between points (x_{1}, y_{1}) and (x_{2}, y_{2}) is given by distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} Now, AB = \sqrt{{(1 - 0)}^{2} + {(4 - 1)}^{2}} = \sqrt{1^{2} + 3^{2}} = \sqrt{1 + 9} = \sqrt{10} units Now, BC = \sqrt{{(4 - 1)}^{2} + {(3 - 4)}^{2}} = \sqrt{{(3)}^{2} + {(- 1)}^{2}} = \sqrt{9 + 1} = \sqrt{10} units Now, CD = \sqrt{{(3 - 4)}^{2} + {(0 - 3)}^{2}} = \sqrt{{(- 1)}^{2} + {(- 3)}^{2}} = \sqrt{1 + 9} = \sqrt{10} units Now, DA = \sqrt{{(0 - 3)}^{2} + {(1 - 0)}^{2}} = \sqrt{{(- 3)}^{2} + {(1)}^{2}} = \sqrt{9 + 1} = \sqrt{10} units Now, AC = \sqrt{{(4 - 0)}^{2} + {(3 - 1)}^{2}} = \sqrt{16 + 4} = \sqrt{20} = 2 \sqrt{5} units Now, BD = \sqrt{{(3 - 1)}^{2} + {(0 - 4)}^{2}} = \sqrt{4 + 16} = \sqrt{20} = 2 \sqrt{5} units So, in quadrilateral ABCD, we have, AB = BC = CD = DA and AC = BD So, given quadrilateral is a square . We know that a quadrilateral is a square, if each of its sides are equal and also its diagonals are equal .$

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