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Ex 4.3, 5 - If area of triangle is 35, (2, -6), (5, 4), (k, 4) - Ex 4.3

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Ex 4.3, 5 If area of triangle is 35 square units with vertices (2, −6), (5, 4), and (k, 4). Then k is • 12 B. −2 C. −12, −2 D. 12, −2 The area of triangle is given by ∆ = 1﷮2﷯ x1﷮y1﷮1﷮x2﷮y2﷮1﷮x3﷮y3﷮1﷯﷯ Here, Area of triangle is 35 square unit Since area is always positive , So ∆ can have positive & negative value ⇒ ∆ = ± 35 square unit Also, x1 = 2 , y1 = – 6 x2 = 5 , y2 = 4 x3 = k , y3 = 4 Putting values ± 35 = 1﷮2﷯ 2﷮−6﷮1﷮5﷮4﷮1﷮k﷮4﷮1﷯﷯ ± 35 = 1﷮2﷯ 2 4﷮1﷮4﷮1﷯﷯−(−6) 5﷮1﷮k﷮1﷯﷯+1 5﷮4﷮k﷮4﷯﷯﷯ ± 35 = 1﷮2﷯ (2 (4 – 4) + 6 (5 – k) + 1 (20 – 4k)) ± 35 = 1﷮2﷯ (2(0) + 6 (5 – k) + (20 – 4k) ± 35 = 1﷮2﷯ (50 – 10k) ±35 × 2 = (50 – 10k) ± 70 = (50 – 10k) So, 70 = 50 – 10k or – 70 = 50 – 10k Hence , the required value of k is –2 or 12 So correct answer is D

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