# Ex 4.3, 5 - Chapter 4 Class 12 Determinants

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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 ∆ = 12 x1y11x2y21x3y31 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 = 12 2−61541k41 ± 35 = 12 2 4141−(−6) 51k1+1 54k4 ± 35 = 12 (2 (4 – 4) + 6 (5 – k) + 1 (20 – 4k)) ± 35 = 12 (2(0) + 6 (5 – k) + (20 – 4k) ± 35 = 12 (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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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.