Question
If the points P(2, 3), Q(5, a) and R(6, 7) are
co-linear, then find the value of 'a'.Solution
If 3 given points are collinear, then the area of the triangle is zero. For P(x1, y1), Q(x2, y2) and R(x3, y3), the area of the triangle is So, 1/2 X [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] = 0 For the given set of points, area of the triangle = (1/2) X [2(a - 7) + 5(7 - 3) + 6(3 - a)] = 0 Or, (1/2) X [2a - 14 + 20 + 18 - 6a] = 0 Or, (1/2) X (24 - 4a) = 12 - 2a = 0 So, 2a = 12 And a = (12/2) = 6
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