For what value of k are the points (– 8, 1), (k, – 4) and (2, –

1.	For what value of k are the points (– 8, 1), (k, – 4) and (2, – 5) collinear?
Answer» Let the given points be A(x₃, y₃) = (2, – 5). So, the area of ∆ABC is ∆ = \(\frac{1}{2}\)\|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)\| = \(\frac{1}{2}\)\|8 (−4 + 5) + k(−5 − 1) + 2 (1 + 4)\| = \(\frac{1}{2}\)\|−6k + 2\| = –3k + 1 or 3k – 1 Since A, B, C, are collinear. ∴ ∆ = 0 ⇒ – 3k + 1 = 0 or 3k - 1 = 0 ⇒ k = \(\frac{1}{3}\)

For what value of k are the points (– 8, 1), (k, – 4) and (2, – 5) collinear?

Answer»

Let the given points be A(x₃, y₃) = (2, – 5).

So, the area of ∆ABC is

∆ = \(\frac{1}{2}\)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

= \(\frac{1}{2}\)|8 (−4 + 5) + k(−5 − 1) + 2 (1 + 4)|

= \(\frac{1}{2}\)|−6k + 2|

= –3k + 1 or 3k – 1

Since A, B, C, are collinear.

∴ ∆ = 0

⇒ – 3k + 1 = 0 or 3k - 1 = 0

⇒ k = \(\frac{1}{3}\)