The points A (1, 2, -1), B (5, -2, 1), C (8, -7, 4), D (4, -3, 2) form_____________(a) trapezium(b) rhombus(c) square(d) parallelogramI have been asked this question in an international level competition.This question is from Three Dimensional Geometry topic in chapter Three Dimensional Geometry of Mathematics – Class 11

The points A (1, 2, -1), B (5, -2, 1), C (8, -7, 4), D (4, -3, 2) form

1.	The points A (1, 2, -1), B (5, -2, 1), C (8, -7, 4), D (4, -3, 2) form_____________(a) trapezium(b) rhombus(c) square(d) parallelogramI have been asked this question in an international level competition.This question is from Three Dimensional Geometry topic in chapter Three Dimensional Geometry of Mathematics – Class 11
Answer» The correct ANSWER is (d) parallelogram The best explanation: We know, DISTANCE between two POINTS (x1, y1, z1) and (x2, Y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\). Distance AB = \(\sqrt{(1-5)^2+(2+2)^2+(-1-1)^2}\) = 6 Distance BC = \(\sqrt{(5-8)^2+(-2+7)^2+(1-4)^2} = \sqrt{43}\) Distance CD = \(\sqrt{(8-4)^2+(-7+3)^2+(4-2)^2}\) = 6 Distance AD = \(\sqrt{(1-4)^2+(2+3)^2+(-1-2)^2} = \sqrt{43}\) Since AB=CD and BC=AD i.e. opposite two sides are equal so, it is a parallelogram.

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