find the value of k for which the points (7,-2),(5,1),(3,k) are collinear
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1/2 [x1(y2-y3)+x2 (y3-y1) + x3(y1-y2)] = 0
[7(1-k) + 5(k+2) + 3(-2-1)]=0
7-7k +5k+10-9 = 0
8-2k=0
2k=8
k=8/2 = 4
[7(1-k) + 5(k+2) + 3(-2-1)]=0
7-7k +5k+10-9 = 0
8-2k=0
2k=8
k=8/2 = 4
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