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What will be the value of p, for which the points (-3, p), (-1, 2) and (1, 1) are collinear?(a) p = -4(b) p = 4(c) p = 3(d) p = -3I had been asked this question in unit test.The doubt is from Geometry topic in chapter Coordinate Geometry of Mathematics – Class 10 |
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Answer» RIGHT option is (c) p = 3 The explanation is: We know that, area of triangle = \(\frac {1}{2}\){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 )} The VERTICES of the triangle are (-3, p), (-1, 2) and (1, 1) The area of triangle = \(\frac {1}{2}\) {-3(2 – 1) + (-1)(1 – p) + 1(p – 2)} = \(\frac {1}{2}\) {-3 – 1 + p + p – 2} = \(\frac {-6 + 2p}{2}\) = – 3 + p Since, the points are collinear; the area of triangle will be zero. -3 + p = 0 p = 3 |
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