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 ( 3k − 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 ( x1,y1 ) , ( x2,y2 ) and ( x3,y3 )  = |x1 ( y2 - y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 ) |

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
⇒ 6k2 - 15k -2k + 5 - 2k2 + 5k - 5 = 0

⇒ 6k2−17k + 5−2k2 + 5k−5=0

⇒ 4k2−12k=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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