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Question

For what value of k are the points (k, 2 – 2k), (-k + 1, 2k), (-4 –k, 6 – 2k) collinear?

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Solution


Δ=12∣ ∣k22k1k+12k14k62k1∣ ∣=0
Apply R2R1 and R3R1
Δ=12∣ ∣k22k12k+14k2042k40∣ ∣=0
or 2k+14k242k4=0
or 4(12k)(42k)(4k2)=0
or (12k)+(k+2)(2k1)=0
or 12k+(2k2+3k2)=0
or 2k2+k1=0
or (k+1)(2k1)=0
k=12,1

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