# 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

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