Question

FIND THE VALUE OF P,IF (-2)^p+1. (-2)P =-8​

Answer 1

Answer:

For the points to be collinear , p = 4

Step-by-step explanation:

Given ,

Points = ( p+1 , 2p-2 ), ( p-1 , p) and (p-3 , 2p-6)

For the given points ( x₁ , y₁ ) , ( x₂ , y₂ ) and ( x₃ , y₃) to be collinear then

[ x₁ ( y₂ - y₃ ) + x₂( y₃ - y₁ )+ x₃( y₁- y₂ )] = 0

Here,

x₁  =  p + 1              y₁ = 2p - 2

x₂ =  p - 1               y₂ = p

x₃  = p - 3               y₃ = 2p - 6

Substituting the values in the formula ,

( p + 1 ) ( p - ( 2p - 6 ) ) + ( p - 1 ) ( 2p - 6 - ( 2p - 2 ) ) + ( p - 3 ) ( 2p - 2 - ( p ) ) = 0

( p + 1 ) ( p -  2p + 6 ) ) + ( p - 1 ) ( 2p - 6 -  2p + 2 ) ) + ( p - 3 ) ( 2p - 2 -  p  ) = 0

( p + 1 ) ( -p + 6 )  + ( p - 1 ) ( -4 ) + ( p - 3 ) ( p - 2 ) = 0

- p² - p + 6p + 6 - 4p + 4 + p² - 3p - 2p + 6 = 0

- 4p + 16 = 0

4p = 16

Dividing both the sides by 4

4p / 4 = 16 / 4

p = 4

Hence,

For the points to be collinear , p = 4

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Answer 2

Answer:

p=2

Step-by-step explanation:

Question is (-2)^p+1.(-2)^p=-8

So,

2p+1+p=(-2)^3                                                 {-8 can be written as (-2)^3}

On comparing powers

2p+1=3

p=2