Let us write by calculating what value of K, the points (1, –1), (2, –1) and (K, –1) lie on same straight line.

Let the points be

A = (x₁, y₁) = (1, -1) and

B = (x₂, y₂) = (2, -1) and

C = (x₃, y₃) = (K, -1)

These points lie on a straight line which means points A, B and C are collinear

As these points are collinear the area of triangle formed by these points would be 0

⇒ Area = 0

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

⇒ × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0

⇒ × [1(-1 – (-1)) + 2(-1 – (-1)) + K(-1 – (-1))] = 0

⇒ × [1(-1 + 1) + 2(-1 + 1) + K(-1 + 1)] = 0

⇒ K × 0 = 0 … (a)

Here we can put any value for K from negative infinity to infinity and our equation (a) is satisfied

Hence K can take any value for the points (1, –1), (2, –1) and (K, –1) to lie on same straight line.

Method 2

If we plot the given points we can observe that K can take any value since the y-coordinate for all the points A, B and C is the same