Question

9. If the points (a, 1), (1, 2) and (0, b+1) are collinear, then show that a b
1 1 + = 1.

Answer 1

Points (a , 1) , (1, 2) and (0, b + 1) are collinear . So area of triangle formed by these points equal to zero.
e.g., area of triangle = 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
⇒0 = 1/2[a(2 - b - 1) + 1(b + 1 - 1) + 0(1 - 2)]
⇒0 = a(1 - b) + b + 0
⇒0 = a - ab + b
⇒a + b = ab
Dividing both sides with ab
⇒a/ab + b/ab = ab/ab
⇒1/b + 1/a = 1 [hence proved ]

Answer 2

Solution :

Let A(a,1) , B(1,2) and C(0,b+1)

are collinear points .

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Let A(x1,y1),B(x2,y2),C(x3,y3)

are collinear then

Area of ∆ABC = 0

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

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Here ,

A( a,1 ) = (x1,y1) ,

B( 1,2) = ( x2,y2 ),

C( 0 , b+1 ) = ( x3,y3)

Area of ∆ABC = 0

=>|a[ 2-(b+1)]+1[b+1-1]+0[1-2]|= 0

=> | a(2-b-1)+b | = 0

=> | a ( 1 - b ) + b | = 0

=> a - ab + b = 0

=> a + b = ab

Divide each term with ab,

We get,

a/ab + b/ab = ab/ab

=> 1/a + 1/b = 1

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