Check sibling questions

Ex 4.3, 3 - Find values of k if area of triangle is 4, vertices

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

 

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Ex 4.3, 3 Find values of k if area of triangle is 4 square units and vertices are (k, 0), (4, 0), (0, 2) The area of triangle is given by ∆ = 1/2 |■8(x1&y1&1@x2&y2&1@x3&y3&1)| Here Area of triangle 4 square units Since area is always positive, ∆ can have both positive & negative signs ∴ Δ = ± 4. Putting x1 = k, y1 = 0, x2 = 4, y2 = 0, x3 = 0 y3 = 2 ±4 = 1/2 |■8(k&0&1@4&0&1@0&2&1)| ±4 = 1/2 |■8(k&0&1@4&0&1@0&2&1)| ± 4 = 1/2 (k|■8(0&1@2&1)|−0|■8(4&1@0&1)|+1|■8(4&0@0&2)|) ± 4 = 1/2 [ k (0 – 2) –0 (4 – 0) + 1 (8 – 0) ] ± 4 × 2 = (k (–2) – 0 + 1 (8)) ± 8 = –2k + 8 So 8 = –2k + 8 or – 8 = –2k + 8 Solving 8 = −2k + 8 8 – 8 = –2k 0 = –2k k = 0/(−2) = 0 +Solving –8 = –2k + 8 –8 – 8 = –2k –16 = –2k k = (−16)/(−2) = 8 So, the required value of k is k = 8 or k = 0

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.