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−(2−2k)1−k−k=6−2k−2k−k−4−1+k 4k−21−2k=6−4k−5 120k+10=6−4k−12k+8k2 8k2+4k−4=0 2k2+k−1=0 2k2(k+1)−1(k+1)=0 (2k−1)(k+1)=0 2k-1=0 and k+1=0 k=12andk=−1 we can also say, slope of AB = slope of AC 2k−(2−2k)1−k−k=6−2k−(2−2k)−k−4−k 4k−21−2k=4−2k−4 −8k2+4k−16k+8=4−8k 8k2+4k−4=0 2k2+k−1=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.