# Ex 4.3,1 - Chapter 4 Class 12 Determinants

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 4.3,1 Find area of the triangle with vertices at the point given in each of the following: • (1, 0), (6, 0), (4, 3) The area of triangle is given by ∆ = 12 x1y11x2y21x3y31 Here, x1 = 1 , y1 = 0 x2 = 6 ,y2 = 0 x3 = 4 ,y3 = 3 ∆ = 12 101601431 = 12 1 0131−0 6141+1 6043 = 12 (1(0 – 3) – 0(6 – 4) + 1 (18 – 0)) = 12 (1( – 3) + 0 + 1 (18) ) = 12 [ – 3 + 18 ] = 152 Thus, the required area of triangle is 𝟏𝟓𝟐 square units Ex 4.3,1 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 ∆ = 12 x1y11x2y21x3y31 Here, x1 = 2 , y1 = 7 x2 = 1 , y2 = 1 x3 = 10 , y3 = 8 ∆ = 12 2711111081 = 12 2 1181−7 11101+1 11108 = 12 ( 2(1 – 8) – 7(1 – 10) + 1 (8 – 10) ) = 12 (2( – 7) – 7( – 9) + 1 (– 2) ) = 12 ( – 14 + 63 – 2 ) = 12 [– 16 + 63] = 472 Thus, the required area of triangle is 𝟒𝟕𝟐 square units Ex 4.3,1 Find area of the triangle with vertices at the point given in each of the following: (iii) (−2, −3), (3, 2), (−1, −8) The area of triangle is given by ∆ = 12 x1y11x2y21x3y31 Here, x1 = – 2, y1 = – 3 x2 = 3 , y2 = 2 x3 = – 1, y3 = – 8 ∆ = 12 −2−31321−1−81 = 12 −2 21−81−(−3) 31−11+1 32−1−8 = 12 [ – 2(2 – ( – 8)) + 3(3 – ( – 1)) + 1 ( – 24 – ( –2) ] = 12 [ – 2(2 + 8) + 3(3 + 1) + 1 (– 24 + 2) ] = 12 [ – 2(10) + 3 (4) + 1 ( – 22) ] = 12 ( – 20 – 22 + 12) = 12 (− 30) = – 15 Since the area of triangle is always positive, the required area of triangle is 15 square units

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.