If (1, 5), (p, 1) and (4, 11) are collinear, find the value of p.
If 3 pts are collinear, then ar(ABC)= 0
A= (1,5) B=(p,1) C=(4,11)
ar(ABC) = 1/2 [x1(y2-y3)+x2(y3-y1) + x3 (y1-y1) ]
0 = [ 1(1-11) + p (11-5) + 4 (5-1) ]
0 = [ -10 +6p + 16 ]
0 = 6p + 6
6p = -6
p = -1