# The points (p,2),(-3,4) and (-7,1) are collinear then find the value of P

2
by zk86012
ok
yes..just wait a little while
giving..?
hope you understood

2015-03-31T23:28:23+05:30
If the 3 points are collinear then the area formed by the triangle of the given points shall be 0
A(p,2)    B(-3,4)   C(-7,1)

So, Area of ABC =1/2 [x1 (y2 - y3) + x2(y3 - y1) + x3 (y1 - y2)] = 0

substituting the values,

=> 1/2{p (2) -3 (-1) -7 (-2)} =0

=> 2p +3 +14 =0
=> 2p = -17
=> p =-17/2

I think -3(-1) is equal to +3
2015-04-01T06:49:31+05:30

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If 3 points are collinear then Ιx₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)Ι=0
x₁=p,y₁=2,x₂=-3,y₂=4,x₃=-7,y₃=1.
Ιp(4-1)+(-3)(1-2)+(-7)(2-4)Ι=0
Ι3p+3+14Ι=0
3p=-17
p=-17/3

Hope this helps you ............... :)
then why was it reported by someone else why do they do so
i don't understand
i think he reported urs bcoz u missed to keep 1/2 in the formula. but although u keep the result will be same :)
oh may be
yaah :)