Question

find the value of k for which the points are collinear {7,-2} , {5,1} , {3,k}.​

Answer 1

Answer

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Answer 2

we know that area of ∆ ABC= 1/2 [x1(y3-y3)+x2(y3-y1)+x3(y1-y2)]

1/2 [7(1-k)+5(k+2)+3(-2-1)]=0

1/2[7-2k+1]=0 (by solving above)

1/2(8-2k)=0

8-2k=0 or 2k =8

k=8/2

k=4