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Mathematics

Name the type of quadrilateral formed, if any, 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).

By formula,

Distance between two points (D) = (x2−x1)2+(y2−y1)2\sqrt{(x2 - x1)^2 + (y2 - y1)^2}

Calculating sides, we get :

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

Calculating diagonals, we get :

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

Since, AB = BC = CD = AD and AC = BD.

Hence, the above points form a square.

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