Q. 234.2( 11 Votes )

# Find the value (s) of k for which the points (3k - 1, k - 2), (k, k - 7) and (k - 1,-k - 2) are collinear.

Answer :

Let A ( 3*k* − 1, *k* − 2 ) , B ( *k*, *k* − 7 ) and C ( *k* − 1, −*k* − 2 ) be the given points.

For points to be collinear area of triangle formed by the vertices must be zero.

Area of the triangle having vertices ( x_{1},y_{1} ) , ( x_{2},y_{2} ) and ( x_{3},y_{3} ) = |x_{1} ( y_{2} - y_{3} ) + x_{2} ( y_{3} - y_{1} ) + x_{3} ( y_{1} - y_{2} ) |

area of ∆ABC = 0

⇒ ( 3k−1 ) [ ( k−7 ) − ( −k−2 ) ] + k [ ( −k−2 ) − ( k−2 ) ] + ( k−1 ) [ ( k−2 ) − ( k−7 ) ] =0

⇒ ( 3k−1 ) [ k−7 + k + 2 ] + k [ −k−2 − k+ 2 ] + ( k−1 ) [ k−2 − k + 7 ] =0

⇒ ( 3k−1 ) ( 2k−5 ) + k (−2k ) + 5 ( k −1 ) =0

⇒ 6k^{2} - 15k -2k + 5 - 2k^{2} + 5k - 5 = 0

⇒ 6k^{2}−17k + 5−2k^{2} + 5k−5=0

⇒ 4k^{2}−12k=0

⇒ 4k ( k−3 ) =0

⇒ k=0 or k−3=0

⇒ k=0 or k=3

Hence, the value of *k* is 0 or 3.

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