Q. 374.3( 9 Votes )

# Find the value (s

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.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Doubt Session47 mins  Section Formula53 mins  Basics of Coordinate Geometry53 mins  Distance Formula52 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 