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Prove that the points A(2, 3), B(-2, 2), C(-1, -2) and D(3, -1) are the vertices of a square ABCD.

Coordinate Geometry

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Answer

By distance formula,

Prove that the points A(2, 3), B(-2, 2), C(-1, -2) and D(3, -1) are the vertices of a square ABCD. Coordinate Geometry, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

d = (x2−x1)2+(y2−y1)2\sqrt{(x2 - x1)^2 + (y2 - y1)^2}

AB=(−2−2)2+(2−3)2=(−4)2+(−1)2=16+1=17.BC=[−1−(−2)]2+[−2−2]2=[−1+2]2+[−4]2=12+16=17.CD=[3−(−1)]2+[−1−(−2)]2=[3+1]2+[−1+2]2=42+12=16+1=17.AD=(3−2)2+(−1−3)2=12+(−4)2=1+16=17.AB = \sqrt{(-2 - 2)^2 + (2 - 3)^2} \\[1em] = \sqrt{(-4)^2 + (-1)^2} \\[1em] = \sqrt{16 + 1} \\[1em] = \sqrt{17}. \\[1em] BC = \sqrt{[-1 - (-2)]^2 + [-2 - 2]^2} \\[1em] = \sqrt{[-1 + 2]^2 + [-4]^2} \\[1em] = \sqrt{1^2 + 16} \\[1em] = \sqrt{17}. \\[1em] CD = \sqrt{[3 - (-1)]^2 + [-1 - (-2)]^2} \\[1em] = \sqrt{[3 + 1]^2 + [-1 + 2]^2} \\[1em] = \sqrt{4^2 + 1^2} \\[1em] = \sqrt{16 + 1} \\[1em] = \sqrt{17}. \\[1em] AD = \sqrt{(3 - 2)^2 + (-1 - 3)^2} \\[1em] = \sqrt{1^2 + (-4)^2} \\[1em] = \sqrt{1 + 16} \\[1em] = \sqrt{17}.

Calculating diagonals,

AC=(−1−2)2+(−2−3)2=(−3)2+(−5)2=9+25=34.BD=[3−(−2)]2+(−1−2)2=[3+2]2+(−3)2=52+9=25+9=34.AC = \sqrt{(-1 - 2)^2 + (-2 - 3)^2} \\[1em] = \sqrt{(-3)^2 + (-5)^2} \\[1em] = \sqrt{9 + 25} \\[1em] = \sqrt{34}. \\[1em] BD = \sqrt{[3 - (-2)]^2 + (-1 - 2)^2} \\[1em] = \sqrt{[3 + 2]^2 + (-3)^2} \\[1em] = \sqrt{5^2 + 9} \\[1em] = \sqrt{25 + 9} \\[1em] = \sqrt{34}.

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

Hence, proved that A, B, C and D are the vertices of a square.

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