Question

find the value k so that the area of triangle ABC with A(k+1,1),B(4,-3),C(7,-k)is 6 sq unit​

Answer 1

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Here,

x₁ = k+1  ; y₁ = 1

x₂ = 4    ; y₂ = -3

x₃ = 7    ; y₃ = -k

We know that,

Area of Δ = 1/2([(x₁y₂ + x₂y₃ + x₃y₁) - (y₁x₂ + y₂x₃ + y₃x₁)]

6 = 1/2[(-3(k+1) - 4k + 7) - (4 - 21 - k(k+1))]

6 × 2 = [(-3k - 3 - 4k +7) - (4 - 21 - k² -k)]

12 = [4 - 7k  -(-17 - k² -k )]

12 = [4 - 7k  + 17 + k² + k]

12 = 21 - 6k + k²

k² - 6k + 21 - 12 = 0

k² - 6k + 9 = 0

k² - 3k - 3k +  9 = 0

k(k - 3) -3( k - 3)  = 0

(k - 3)(k - 3) = 0

(k - 3)² = 0

k - 3 = 0

k = 3