(Q23) Find the value of 'K' for which the points are collinear (7,-2) , (5,1) , (3,K).

Given : A(7,-2) , B(5,1) and C(3,K) are collinear.

Area of ∆ABC = 0

But area of triangle,

=
1
2
 |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
=
1
2
 |7( - K)+5(K + )+3(- - )= 0|

= | - 7k + 5k + - |= 0

= |-2k + |= 0

= -2k + = 0

-2k = -

K=
2

K =