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  • In each of the following find the value of ‘k’, for which the points are collinear. (8, 1), (k, – 4), (2, –5)

2.(ii) In each of the following find the value of ‘k’, for which the points are collinear.  (8, 1), (k, – 4), (2, –5)

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The points (8,1), (k, -4), (2,-5) are collinear if the area of the triangle formed by these 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 [ 8(-4-(-5))+k((-5)-1)+2(1-(-4)) \right ] = 0

\Rightarrow 8-6k+10 = 0

\Rightarrow 6k = 18

\Rightarrow k = 3

Hence, the points are collinear for k = 3.

Posted by

Divya Prakash Singh

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