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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)

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)

Posted by

Divya Prakash Singh

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