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find the value of k, if the points A(2,3) B(4,k) C(6,-3) are collinear.

The given points A 2, 3,  B 4, K and C 6, -3 are collinear if area of the △ ABC formed by joining these points = 0or   area △ ABC = 0⇒ 12 x1 y2 - y3 + x2  y3 - y1 + x3 y1 - y2 = 0⇒ 12 2 K + 3 + 4 -3 - 3 + 6 3 - K = 0⇒ 12 2K + 6 - 24 + 18 - 6K = 0⇒ 12 -4K = 0⇒ -4K = 0⇒ K = 0

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AREA=0=1/2{2(K+3)+4(-3+2) +6(3-K)}

2K+6-4+18-6K=0

--4K+20=0

4K=20

K=5

HOPE THIS HELPS U

  • -4

thnxx suril :)

  • -8

How does 4(-3+2) come???

  • -5

K should be 0

  • -5

how 4(-3+2) comes when its clearly mention that y3-y1? and y3 is -3 and y1 is 3?

  • -4

correct Abhishek

  • -4

I got K as 0

  • -4

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

2(K+3)+4(-3-3)+6(3-K) = 0

2k+6+4(-6)+18-6k = 0

2k+6-24+18-6k = 0

2k+24-24-6k = 0

-4k = 0

k=0

hope it helps.

  • 26
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