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

Ex 4.3, 5 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.3, 5 - Chapter 4 Class 12 Determinants - Part 3

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Ex 4.2, 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&[email protected]&y2&[email protected]&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&[email protected]&4&[email protected]&4&1)| ±35 = 1/2 (2|■8(4&[email protected]&1)|−(−6)|■8(5&[email protected]&1)|+1|■8(5&[email protected]&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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.