Example 14 - Chapter 7 Class 10 Coordinate Geometry

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Example 14 Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear. If the above points are collinear, they will lie on the same line, i.e. they will not form triangle Therefore, Area of ∆ABC = 0 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] = 0 Here x1 = 2 , y1 = 3 x2 = 4 , y2 = k x3 = 6 , y3 = −3 Putting values 1/2 [ 2(k – (−3)) + 4(−3 − 3) + 6(3 – k) ] = 0 2(k + 3) + 4(−6) + 6(3 − k) = 0 × 2 2k + 6 – 24 + 18 – 6k = 0 2k – 6k = − 6 + 24 −18 −4k = 0 ⇒ k = 0