Given: – Points are (k, 2 – 2 k), (– k + 1, 2k) and (– 4 – k, 6 – 2k) which are collinear

Tip: – For Three points to be collinear, the area of the triangle formed by these points will be zero.

If vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

Expanding along R₁

⇒

⇒ k(2k – 6 + 2k) – (2 – 2k)(– k + 1 + 4 + k) + 1(6 – 2k – 6k + 2k² + 8k + 2k²) = 0

⇒ 4k² – 6k – 10 + 10k + 6 + 4k² = 0

⇒ 8k² + 4k – 4 = 0

⇒ 8k² + 8k – 4k – 4 = 0

⇒ 8k(k + 1) – 4(k + 1) = 0

⇒ (8k – 4)(k + 1) = 0

If 8k – 4 = 0

⇒

And, If k + 1 = 0

⇒ K = – 1

Hence, k = – 1, 0.5