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Ex 4.3,1 - Chapter 4 Class 12 Determinants - Part 5

Ex 4.3,1 - Chapter 4 Class 12 Determinants - Part 6

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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 ∆ = 1﷮2﷯ x1﷮y1﷮1﷮x2﷮y2﷮1﷮x3﷮y3﷮1﷯﷯ Here, x1 = – 2, y1 = – 3 x2 = 3 , y2 = 2 x3 = – 1, y3 = – 8 ∆ = 1﷮2﷯ −2﷮−3﷮1﷮3﷮2﷮1﷮−1﷮−8﷮1﷯﷯ = 1﷮2﷯ −2 2﷮1﷮−8﷮1﷯﷯−(−3) 3﷮1﷮−1﷮1﷯﷯+1 3﷮2﷮−1﷮−8﷯﷯﷯ = 1﷮2﷯ [ – 2(2 – ( – 8)) + 3(3 – ( – 1)) + 1 ( – 24 – ( –2) ] = 1﷮2﷯ [ – 2(2 + 8) + 3(3 + 1) + 1 (– 24 + 2) ] = 1﷮2﷯ [ – 2(10) + 3 (4) + 1 ( – 22) ] = 1﷮2﷯ ( – 20 – 22 + 12) = 1﷮2﷯ (− 30) = – 15 Since the area of triangle is always positive, the required area of triangle is 15 square units

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