for what value of p are the points (-3,9),(2,p)and(4,-5) collinear
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Answered by
8
Answer:
x₁ = -3
x₂ = 2
x₃ = 4
y₁ = 9
y₂ = p
y₃ = -5
For the points to be collinear :-
x₁(y₂ -y₃) + x₂(y₃ - y₂) + x₃(y₁ - y₂) = 0
-3( p - (-5)) + 2(-5 - 9) + 4( 9 - p) = 0
-3( p + 5) + 2(-14) + 4( 9 - p) = 0
-3p - 15 - 28 + 36 - 4p = 0
-7p - 7 = 0
-7p = 7
p = 7/-7
p = -1
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Answered by
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Let the points be A (-3,9),B (2,p) and C(4,-5)
If A,B,C are collinear
Ar (ABC)=0
THEREFORE PUTTING FORMULA
ar (ABC)=1/2 [ X1 (Y2-Y3) + X2 (Y3-Y1) +X3 (Y1-Y2) ]=0
= 1/2 [ -3 {p-(-5)} + 2 { -5-9} + 4 (9-p) ]=0
=1/2 [ -3p-15-28+36-4p ]=0
-7p-7=0
-7=7p
P=-1
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