points are collenear 3k-1 k-2 k k-7 k-1 -k-2 find the value of k
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For collinearity of the points ,area of the triangle formed by given points is zero 1÷2{(3k-1)×(k-7+k+2)+k(-k-2-k+2)+(k-1)(k-2-k+7)}=0{(3k-1)(2k-5)-2k²+5k-5}=04k²-12k=0K=0,3 ok
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