Q. 34.6( 5 Votes )

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

Answer :

Let the points be

A = (x_{1}, y_{1}) = (1, -1) and

B = (x_{2}, y_{2}) = (2, -1) and

C = (x_{3}, y_{3}) = (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 = × [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

Where (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are vertices of triangle

⇒ × [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = 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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