(Q25) Find the value of 'K' for which the points are collinear (K,K) , (2,3) and (4,-1).

Given : A(K,K) , B(2,3) and C(4,-1) are collinear.

Area of ∆ABC = 0.

=
1
2
 |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)= 0|
=
1
2
 |K( + ) + 2(- - K) + 4(K - )| = 0

= |4K - -2K +4K -| = 0

= |6K - | = 0

6K - = 0

6K =

K =
6
 =
3
K =
3