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Find the values of k for which the points A(k + 1, 2k), B(3k, 2k + 3) and (5k – 1, 5k) are collinear.

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#### Solution

A(k + 1, 2k) , B(3k, 2k + 3) and (5k – 1, 5k)

If 3 points are collinear then area of triangle formed by them = 0

`1/2[(k+1)(2k+3-5k)- 2k(3k-5k+1)+1(15k^2- 10k^2- 2k+ 15k - 3)= 0`

`1/2[-3k^2+3k-3k+3+4k^2-2k+15k^2-10k^2-2k+15k-3]=0`

`1/2[6k^2+11k]=0`

`6k^2+11k=0`

`k=0,-11/6`

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