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| 1. |
Find the value of p for which the points (p+1,2p-2) , (p-1,p) , (p-3,2p-6) |
| Answer» For the points to be collinear , p = 4Step-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 )] = 0Here,x₁ = p + 1 y₁ = 2p - 2x₂ = p - 1 y₂ = px₃ = p - 3 y₃ = 2p - 6Substituting 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 = 04p = 16Dividing both the sides by 44p / 4 = 16 / 4p = 4Hence,For the points to be collinear , p = 4 | |