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  1. Chapter 4 Class 12 Determinants
  2. Concept wise

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 ∆ = 1/2 |■8(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 |■8(2&−6&1@5&4&1@k&4&1)| ±35 = 1/2 (2|■8(4&1@4&1)|−(−6)|■8(5&1@k&1)|+1|■8(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 Solving 70 = 50 – 10k 70 – 50 = –10k 20 = – 10k –10k = 20 k = 20/(−10) = – 2 Solving –70 = 50 – 10k –70 – 50 = –10k –120 = – 10k –10k = –120 k = (−120)/(−10) = 12

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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.