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

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The points (7, –2), (5, 1), (3, k) are collinear if the area of the triangle formed by the points will be zero.

Area of the triangle is given by:

Area = \frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ] = 0

Substituting the values in the above equation, we obtain

\frac{1}{2}\left [ 7(1-k)+5(k-(-2))+3(-2-1) \right ] = 0

\left [ 7-7k+5k+10-9 \right ] = 0

\Rightarrow -2k+8 = 0

\Rightarrow k = 4

Hence, the points are collinear for k=4.

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Divya Prakash Singh

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

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Divya Prakash Singh

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