InterviewSolution
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InterviewSolution
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Show that the points A (1,- 2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram. |
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Answer» Vertices of a parallelogram ABCD are: A (1,- 2), B (3, 6), C (5, 10) and D (3, 2) Length of side AB = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) Length of side AB = \(\sqrt{(3 - 1)^2 + (6 + 2)^2}\) = \(\sqrt(4 + 64)\) = \(\sqrt68\) units Length of side BC = \(\sqrt{(5 - 3)^2 + (10 - 6)^2}\) = \(\sqrt(4 + 16)\) = \(\sqrt20\) units Length of side CD = \(\sqrt{(3 - 5)^2 + (2 - 10)^2}\) = √(4 + 64) = √68 units Length of side DA = \(\sqrt{(3 - 1)^2 + (2 + 2)^2}\) = √(4+16) = √20 units Length of diagonal BD = \(\sqrt{(3 - 3)^2 + (2 - 6)^2}\) = √16 = 4 units Length of diagonal AC = \(\sqrt{(5 - 1)^2 + (10 + 2)^2}\) = √(16+144) = √160 units Opposite sides of the quadrilateral formed by the given four points are equal i.e. (AB = CD) & (DA = BC)Also, the diagonals BD & AC are unequal.Therefore, the given points form a parallelogram. |
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