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Question

If the points (k,2-2k), (1-k, 2k) and (-k-4, 6-2k) are collinear, possible values of k are ..............

A

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B

1/2

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C

1

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D

-1

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Solution

The correct options are
B

1/2


D

-1


Points A(k,2-2k), B (1-k, 2k) and C(-k-4, 6-2k) are collinear.
Then, slope of line AB = slope of line BC
2k(22k)1kk=62k2kk41+k
4k212k=64k5
120k+10=64k12k+8k2
8k2+4k4=0
2k2+k1=0
2k2(k+1)1(k+1)=0
(2k1)(k+1)=0
2k-1=0 and k+1=0
k=12 and k=1
we can also say, slope of AB = slope of AC 2k(22k)1kk=62k(22k)k4k
4k212k=42k4
8k2+4k16k+8=48k
8k2+4k4=0
2k2+k1=0
we got the same equation of previous case
So k=12 and k=-1 are the only possible values of k when given points are collinear.

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