Ex 4.2,1 (ii) - Chapter 4 Class 12 Determinants
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 4.2, 1 Find area of the triangle with vertices at the point given in each of the following: (ii) (2, 7), (1, 1), (10, 8) The area of triangle is given by ∆ = 1/2 |■8(x1&y1&1@x2&y2&1@x3&y3&1)| Here, x1 = 2 , y1 = 7 x2 = 1 , y2 = 1 x3 = 10 , y3 = 8 ∆ = 1/2 |■8(2&7&1@1&1&1@10&8&1)| = 1/2 (2|■8(1&1@8&1)|−7|■8(1&1@10&1)|+1|■8(1&1@10&8)|) = 1/2 ( 2(1 – 8) – 7(1 – 10) + 1 (8 – 10) ) = 1/2 (2(–7) – 7(–9) + 1 (–2) ) = 1/2 ( – 14 + 63 – 2 ) = 1/2 [– 16 + 63] = 47/2 Thus, the required area of triangle is 𝟒𝟕/𝟐 square units
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo