The four points whose coordinates are (2, 1), | vedclass

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(C) Let the points be $A(2, 1), B(1, 4), C(4, 5), D(5, 2)$.Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2

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a square

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(C) Let the points be $A(2, 1), B(1, 4), C(4, 5), D(5, 2)$.Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:$AB = \sqrt{(1-2)^2 + (4-1)^2} = \sqrt{(-1)^2 + 3^2} = \sqrt{1+9} = \sqrt{10}$$BC = \sqrt{(4-1)^2 + (5-4)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}$$CD = \sqrt{(5-4)^2 + (2-5)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1+9} = \sqrt{10}$$DA = \sqrt{(2-5)^2 + (1-2)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10}$Since all sides are equal,the quadrilateral is a rhombus.Now,calculate the lengths of the diagonals $AC$ and $BD$:$AC = \sqrt{(4-2)^2 + (5-1)^2} = \sqrt{2^2 + 4^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$$BD = \sqrt{(5-1)^2 + (2-4)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$Since all sides are equal and the diagonals are equal,the quadrilateral is a square.

The four points whose coordinates are $(2, 1), (1, 4), (4, 5), (5, 2)$ form:

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