

Get live Maths 1-on-1 Classs - Class 6 to 12
Examples
Example 2 Important
Example 3 Important
Example 4
Example 5 Important
Example 6
Example 7 Important
Example 8
Example 9 Important
Example 10 Important
Example 11 Deleted for CBSE Board 2023 Exams
Example 12 Deleted for CBSE Board 2023 Exams
Example 13 Important Deleted for CBSE Board 2023 Exams
Example 14 Important Deleted for CBSE Board 2023 Exams You are here
Example 15 Deleted for CBSE Board 2023 Exams
Last updated at March 16, 2023 by Teachoo
Example 14 Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear. If points A, B, C 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 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