(7,2),(5,1),(3,k) points are Collinear. Find the value of k

1.	(7,2),(5,1),(3,k) points are Collinear. Find the value of k
Answer» A(7,-2), B(5,1), C(3,2K)Δ = 1/2[(x₁y₂+x₂y₃+x₃y₁)-(x₂y₁+x₃y₂+x₁y₃)Δ = 1/2[(7+10K-6)-(-10+3+14K)] = 1/2[1+10K+7-14K] = 1/2[8-4K] = 4-2KGIVEN THE POINTS ARE COLLINEARΔ = 04-2K = 02K = 4K = 2 Area of ∆ABC=01/2{x1(y2-y3)+x2(y3-y2)+x3(y1-y2)=01/2{7(1-k)+5(k-2)+3(2-1)=07-7k+5k-10+3=02k=0k=0 Put area of triangle as 0...then write the formula for area of triangle =0 put all the values of x1, x2, x3, y1, y2, y3.....then yu will get d answer

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