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Mathematics

Name the type of quadrilateral formed by the following points and give reasons for your answer :

(-1, -2), (1, 0), (-1, 2), (-3, 0).

Coordinate Geometry

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Answer

Let coordinates be A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0).

AB=[1−(−1)]2+[0−(−2)]2=[1+1]2+[0+2]2=22+22=4+4=8.BC=(−1−1)2+(2−0)2=(−2)2+(2)2=4+4=8.CD=[−3−(−1)]2+[0−2]2=[−3+1]2+[−2]2=[−2]2+[−2]2=4+4=8.AD=[−3−(−1)]2+[0−(−2)]2=[−3+1]2+[2]2=[−2]2+[2]2=4+4=8.AB = \sqrt{[1 - (-1)]^2 + [0 - (-2)]^2} \\[1em] = \sqrt{[1 + 1]^2 + [0 + 2]^2}\\[1em] = \sqrt{2^2 + 2^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8}. \\[1em] BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} \\[1em] = \sqrt{(-2)^2 + (2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8}. \\[1em] CD = \sqrt{[-3 - (-1)]^2 + [0 - 2]^2} \\[1em] = \sqrt{[-3 + 1]^2 + [-2]^2} \\[1em] = \sqrt{[-2]^2 + [-2]^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8}. \\[1em] AD = \sqrt{[-3 - (-1)]^2 + [0 - (-2)]^2} \\[1em] = \sqrt{[-3 + 1]^2 + [2]^2} \\[1em] =\sqrt{[-2]^2 + [2]^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8}.

Since, AB = BC = CD = AD.

So, ABCD can be square or rhombus.

Calculating diagonal,

AC=[−1−(−1)]2+[2−(−2)]2=[−1+1]2+[2+2]2=0+42=16=4.BD=(−3−1)2+(0−0)2=(−4)2+02=16=4.AC = \sqrt{[-1 - (-1)]^2 + [2 - (-2)]^2} \\[1em] = \sqrt{[-1 + 1]^2 + [2 + 2]^2} \\[1em] = \sqrt{0 + 4^2} \\[1em] = \sqrt{16} \\[1em] = 4. \\[1em] BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} \\[1em] = \sqrt{(-4)^2 + 0^2} \\[1em] = \sqrt{16} \\[1em] = 4.

Since, diagonals are also equal.

Hence, (-1, -2), (1, 0), (-1, 2), (-3, 0) are the vertices of a square.

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