# Ex 4.3,1 (ii) - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Aug. 18, 2021 by Teachoo

Last updated at Aug. 18, 2021 by Teachoo

Transcript

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

Chapter 4 Class 12 Determinants (Term 1)

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.