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

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