Gujarat BoardEnglish MediumSTD 11 ScienceMATHSSTRAIGHT LINES2 Marks
Question
Without using distance formula, show that the points (-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.
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Answer
Let A(-2, -1), B (4, 0), C (3, 3) and D(-3, 2) be vertices of a quadrilateral ABCD. $\therefore$ Slope of AB $ = \frac{{0 - ( - 1)}}{{4 - ( - 2)}} = \frac{1}{6}$ Slope of BC $= \frac{{3 - 0}}{{3 - 4}} = \frac{3}{{ - 1}} = - 3$ Slope of DC $= \frac{{3 - 2}}{{3 - ( - 3)}} = \frac{1}{6}$ Slope of AD $= \frac{{2 - ( - 1)}}{{ - 3 - ( - 2)}} = \frac{3}{{ - 1}} = - 3$ $\therefore$ Slope of AB = Slope of DC $\Rightarrow$ AB|| DC And Slope of BC = Slope of AD $\Rightarrow$ BC || AD Thus ABCD is a parallelogram
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