Find the value of k if points (k, 3), (6, - 2) and (-3, 4) are collinear.
The three given points are A(k, 3), B(6, −2) and C(−3, 4). It is also said that they are collinear and hence the area enclosed by them should be 0.
Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3)
= 
|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
Given that area of ∆ABC = 0
∴ 0 = 
|k(-2 – 4) + 6(4 - 3) - 3(3 – (-2))|
∴ 0 = 
|-6k + 6 - 15|
∴ -
|-6k + 9|= 0
6k + 9 = 0
∴ k = 
Hence, the value of k is 
15