In each of the following find the value of ‘k’, for which the points are collinear:
(i) (7, –2), (5, 1), (3, k) 

(ii) (8, 1), (k, – 4), (2, –5)

(i) Let A(x₁, y₁) = (7, - 2), B(x₂, y₂) = (5 , 1) and C(x₃, y₃) = (3, k)

For three points to be collinear, the area of triangle must be equal to zero.

1/2 [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)] = 0  

By substituting the values of vertices, A, B, C in the above formula,

1/2 [7{1 - k} + 5{k - (- 2)} + 3{(- 2) - 1}] = 0

7 - 7k + 5k + 10 - 9 = 0

- 2k + 8 = 0

k = 4

Hence, the given points are collinear for k = 4

(ii) Let A(x₁, y₁) = (8, 1), B(x₂, y₂) = (k , - 4) and C(x₃, y₃) = (2, - 5)

For three points to be collinear, the area of triangle must be equal to zero.

1/2 [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)] = 0  

By substituting the values of vertices, A, B, C in the above formula,

1/2 [8{- 4 - (- 5)} + k{(- 5) - (1)} + 2{1 - ( - 4)}] = 0

8 - 6k + 10 = 0 (By Transposing)

6k = 18

k = 3

Hence, the given points are collinear for k = 3