Question

for what value of p are the points (-3,9),(2,p)and(4,-5) collinear​

Answer 1

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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Answer 2

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