Question

4. In each of the following, find the value of k for which the given points are collinear.
(i) (k, -1), (2, 1) and (4, 5) (ii) (2,-5),(3,-4)AND(9,K)(iii) (K,K),(2,3)and (4,-1)

Answer 1

Hi ,

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

are the vertices of a ∆ABC , then

Area of the ∆ABC

∆ =1/2|x1 (y2-y3 )+x2 (y3-y1)+x3 (y1-y2)|

Note : If A , B , C three points are

Collinear then area ∆ = 0

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1 ) Let A(x1,y1 ) = ( k , - 1 )

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

C( x3 , y3 ) = ( 4 , 5 )

Area of ∆ABC = 0

Since A, B , C are collinear .

1/2|k(1-5)+2[5-(-1)]+4(-1-1)| = 0

-4k + 2 × 6 + 4 × ( -2 ) = 0

-4k + 12 - 8 = 0

-4k + 4 = 0

-4k = -4

k = ( -4 )/( -4 )

k = 1

2 ) Let A = ( 2 , - 5 )

B = ( 3 , - 4 )

C = ( 9 , k )

Area of ∆ABC = 0

1/2|2(-4-k)+3(k+5)+9(-5+4)| = 0

-8 - 2k + 3k + 15 - 9 = 0

k - 17 + 15 = 0

k = 2

3 ) Let A = ( k , k )

B = ( 2 , 3 )

C = ( 4 , -1 )

Area ∆ ABC = 0

1/2| k( 3 + 1) + 2( -1 - k ) + 4( k - 3 ) | = 0

4k - 2 - 2k + 4k - 12 = 0

6k - 14 = 0

6k = 14

k = 14/6

k = 7/3

I hope this helps you.

: )