(16,-18) ,(3 ,k) ,(-10,6) are collinear find k
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A(16,-18),B(3,k)and C(-10,6)
since A, B and C are collinear then area of triangle ABC=0
so, 16(k-6)+3(6+18)-10(-18-k) =0
16K -96+18+54+180+10k=0
26k-78+234=0
26k+156=0
k=-156/26
k=-6
since A, B and C are collinear then area of triangle ABC=0
so, 16(k-6)+3(6+18)-10(-18-k) =0
16K -96+18+54+180+10k=0
26k-78+234=0
26k+156=0
k=-156/26
k=-6
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