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prove that the points (-7 -3) (5 10) (15 8)and(3 -5) taken in order are the vertices of parallelogram |
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Answer» 8) and (3, -5). Distance between (-7, -3), (5, 10) = √[(5 + 7)2 + (10 + 3)2] = √(144 + 169) = √313 Distance between (5, 10), (15, 8) = √[(15 - 5)2 + (8 - 10)2] = √(100 + 4) = √104 Distance between (15, 8), (3, -5) = √[(3 - 15)2 + (-5 - 8)2] = √(144 + 169) = √313 Distance between (3, -5), (-7, -3) = √[(-7 - 3)2 + (-3 + 5)2] = √(100 + 4) = √104Opposite sides of the quadrilateral FORMED by the given points are equal. So, the given points form a parallelogram.Diagonal of a parallelogram divides it into TWO congruent triangles. So, AREA of parallelogram = 2[Area of the triangle formed by (-7, -3), (5, 10) and (15, 8). = 2[1/2 |-7(10 - 8) + 5(8 + 3) + 15(-3 - 10)|] = |-7(2) + 5(11) + 15(-13)| = |-14 + 55 - 195| = |- 154| = 154 sq units. |
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