Question

If the three points (k, 2k), (2k, 3k) and (3, 1) are collinear then k is equal to

(a) –2 

(b) \(-\frac{1}{2}\)

(c) \(\frac{1}{2}\)

(d) \(\frac{3}{2}\)

Answer 1

(a) - 2

For three points to be collinear, area of the triangle formed by the three points should be equal to zero, i.e.

\(\frac{1}{2}\) [k(3k – 1) + 2k(1 – 2k) + 3(2k – 3k)] = 0

⇒ \(\frac{1}{2}\) [3k² – k + 2k – 4k² – 3k] = 0

⇒ k² + 2k = 0 ⇒ k = 0 or –2 

Neglecting k = 0, as then (k, 2k) and (2k, 3k) will be the same point, we take k = –2.