find the value of k for which the points are colinear ( 3k-1, k-2) , ( k , k-7) and (k-1, -k-2)
 

Dear Student,

Please find below the solution to the asked query:

We have three points   ( 3 k - 1 , k - 2) , ( k , k - 7) and ( k - 1 , - k - 2)

We know if these points are collinear then the area of triangle from given points  is zero , Then given points are collinear  .

We know area of triangle from given three points  :

Area  = 12x1y2 - y3  + x2y3 - y1  + x3y1 - y2 

Here x1 = 3k - 1 , x2 =  k , x3 = k - 1  and  y1 = k - 2 , y2 = k - 7 , y3 = - k - 2

So,

12  3 k - 1   k - 7 -   - k - 2 + k   - k - 2 -   k - 2 + k - 1   k - 2 -   k - 7 =0  3 k - 1   k - 7 -   - k - 2 + k   - k - 2 -   k - 2 + k - 1   k - 2 -   k - 7 =0  3 k - 1    k - 7 + k + 2 + k    - k - 2 -    k + 2 + k - 1    k - 2 -  k + 7 =0  3 k - 1    2 k - 5 + k    -2 k  + k - 1  5 =0   6 k2 - 15 k - 2 k + 5  +- 2 k2 +  5 k - 5 =0 6 k2 - 15 k - 2 k + 5 - 2 k2 + 5 k - 5=0 4 k2 - 12 k  =0 4 k k  - 3  =0 So, k = 0  and  3                                  ( Ans )


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