Question

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

Answer 1

½ [(k + 1)(2k + 3 – 5k) + 3k(5k – 2k) + (5k – 1)(2k – 2k – 3)] = 0

(k + 1)(3 – 3k) + 9k2 – 3(5k – 1) = 0

2k2 – 5k + 2 = 0

(k – 2)(2k – 1) = 0

k = 2, 1/2