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Show that the points A(1, 1), B(-1, -1) and C(-3,3\sqrt{3}, \sqrt{3}) form an equilateral triangle.

Coordinate Geometry

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Answer

By distance formula,

Show that the points A(1, 1), B(-1, -1) and C form an equilateral triangle. 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=(−1−1)2+(−1−1)2=(−2)2+(−2)2=4+4=8=22.BC=[−3−(−1)]2+[3−(−1)]2=[−3+1]2+[3+1]2=3+1−23+3+1+23=8=22.AC=[−3−1]2+[3−1]2=3+1+23+3+1−23=8=22.AB = \sqrt{(-1 - 1)^2 + (-1 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (-2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}. \\[1em] BC = \sqrt{[-\sqrt{3} - (-1)]^2 + [\sqrt{3} - (-1)]^2} \\[1em] = \sqrt{[-\sqrt{3} + 1]^2 + [\sqrt{3} + 1]^2} \\[1em] = \sqrt{3 + 1 - 2\sqrt{3} + 3 + 1 + 2\sqrt{3}} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}. \\[1em] AC = \sqrt{[-\sqrt{3} - 1]^2 + [\sqrt{3} - 1]^2} \\[1em] = \sqrt{3 + 1 + 2\sqrt{3} + 3 + 1 - 2\sqrt{3}} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}. \\[1em]

Since, AB = BC = AC.

Hence, ABC is an equilateral triangle.

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