The points (k,2−2k),(−k+1,2k) and (−4−k,6−2k) are collinear for
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a
All value of k
b
k=−1 or 12
c
k=−1
d
No value of k
answer is C.
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Detailed Solution
The given points are collinear, ifArea of Δ=0⇒ k2−2k1−k+12k1−4−k6−2k1=0Applying R2→R2−R1,R3→R3−R1⇒ −2k+14k−2−4−2k4=0⇒ (1−2k)(k+1)=0⇒ k=−1 or k=12 , neglecting 12 as when k=12 , points are same.