Find the value of K. If the points A(4,6) B(8,2k) and C(12,-6 ) are collinear
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A(4,6)
B(8,2k)
C(12,-6)
Given that A , B and C are collinear
They will lie on a same line i.e. They will not form triangle
AREA OF /\ABC = 0
1/2[x1 ( y2-y3) + x2(y3-y1) + x3(y1-y2)]=0
Here
x1 = 4 , y1 = 6
x2 = 8 , y2 = 2k
x3 =12 , y3 =-6
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=> 1/2[ 4(2k -(-6)+8(-6-6)+ 12(6-2k)] = 0
=> 1/2 [4(2k+6) + 8(-12) + 12(6-2k)] = 0
=> 1/2 [ 8k + 24 - 96 + 72 - 24k] = 0
=> 1/2 [ - 16k + 24 - 24 ] = 0
=> - 16k = 0
=> k = 0
So , value of 'k' is 0..
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