Question
The quadrilateral, whose vertices are A (–1, 1), B
(0, –3), C (5, 2) and D (4, 6) is a:Solution
AB = √ (-1 - 0)2 + (1+3)2 = √ 17 BC= √ (0-5)2 + (-3 -2)2 = √ 50 CD = √ (5-4)2 + (2-6)2 = √ 17 DA = √ (4+1)2 + ( 6 - 1)2 = √ 50 Then AB = CD , and BC = DA Slope m 1 = AC = (y2 – y1)/(x2 – x1) = ( 2 - 1)/(5 + 1) = 1/6 And slope m 2 = BD = (y4 - y3)/(x4 - x3) = (6 + 3)/(4 – 0) = 9/4 Since m1 * m2 ≠ 1 then it is a parallelogram.
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