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Quadrilateral ABCD has vertices A(2;3), B(7;10), C(9;4), and D(4;3). Prove that ABCD is a parallelogram but not a rhombus.

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User Midhu
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Final answer:

To prove ABCD is a parallelogram, we confirm opposite sides are parallel and equal in length. Showing it's not a rhombus involves proving not all sides are of equal length. Using vector and distance calculations confirms ABCD is indeed a parallelogram, but not a rhombus.

Step-by-step explanation:

To prove that ABCD is a parallelogram but not a rhombus, we need to first establish that both pairs of opposite sides are parallel and equal in length. Then, to show it's not a rhombus, we will demonstrate that not all sides are equal in length.

The coordinates of the vertices are A(2,3), B(7,10), C(9,4), and D(4,3). The vector AB is obtained by subtracting the coordinates of A from B, which results in (5,7). Similarly, CD is (5,1), BC is (2,-6), and DA is (2,0). Since AB and CD have the same x-component (indicating they're parallel to the x-axis by the same amount), and BC and DA have the same y-component (indicating they're parallel to the y-axis by the same amount), AB is parallel to CD and BC is parallel to DA.

To check if the lengths of AB and CD are equal, we can use the distance formula. The distance between A and B is √((7-2)² + (10-3)²) = √(25 + 49) = √74. The distance between C and D is √((9-4)² + (4-3)²) = √(25 + 1) = √26. Since the lengths of AB and CD are not equal, we conclude that ABCD is a parallelogram but not a rhombus.

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User Hamy
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