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

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

Gaurav Seth 6 years, 6 months ago

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] = 4-2K

GIVEN THE POINTS ARE COLLINEAR

Δ = 0

4-2K = 0

2K = 4

K = 2

Anushka K 6 years, 6 months ago

Area of ∆ABC=0 1/2{x1(y2-y3)+x2(y3-y2)+x3(y1-y2)=0 1/2{7(1-k)+5(k-2)+3(2-1)=0 7-7k+5k-10+3=0 2k=0 k=0

Gungun_ ❤❤❤ 6 years, 6 months ago

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
http://mycbseguide.com/examin8/

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