Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

Let the points (1, 5), (2, 3), and (- 2, - 11) be represented as A, B, and C.

For A, B, and C to be collinear, they must lie on the same line. 

Hence, we will have to check if AB + BC = AC or BC + AC = AB or AB + AC = BC.

We know that the distance between any two points is given by,

 Distance Formula =   √ [(x₂ ^_-  x₁)² + (y₂ - y₁)²]  ....(1)

To find AB, the Distance between the Points A (1, 5) and B (2, 3), let x₁ = 1, y₁ = 5, x₂ = 2, y₂ = 3

∴ AB = √(2 - 1)² + (3 - 5)²  (By Substituting in (1))

= √5

To find BC Distance between Points B (2, 3) and C (- 2, - 11), let x₁ = 2, y₁ = 3, x₂ = - 2, y₂ = - 11

Therefore, BC = √(-2 - 2)² + (-11 - 3)²

= √(- 4)² + (-14)²  (By Substituting in the Equation (1))

= √16 + 196

= √212

To find AC Distance between Points A (1, 5) and C (-2, -11), let x₁ = 1, y₁ = 5, x₂ = -2, y₂ = -11

Therefore, CA = √(-2 - 1)² + (-11 - 5)²

= √(-3)² + (-16)² (By Substituting in the Equation (1))

= √9 + 256

= √265

AB = √5, BC = √212, CA = √265

Since AB + AC ≠ BC and BC + AC ≠ AB  and AB + BC ≠ AC, therefore, the points (1, 5), (2, 3), and (- 2, - 11) are not collinear.