# Ex 4.3, 3 - Chapter 4 Class 12 Determinants

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

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 ∆ = 12 x1y11x2y21x3y31 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 = 12 k01401021 ± 4 = 12 k 0121−0 4101+1 4102 ± 4 = 12 [ k (0 – 2) – 0 (4 – 0) + 1 (8 – 0) ] ± 4 = 12 ( k ( –2) – 0 + 1 (8)) ± 8 = – 2k + 8 So 8 = – 2k + 8 or – 8 = – 2k + 8 Ex 4.3, 3 Find values of k if area of triangle is 4 square units and vertices are (ii) (-2, 0), (0, 4), (0, k) The area of triangle is given by ∆ = 12 x1y11x2y21x3y31 Here Area of triangle 4 square units Since area is always positive, ∆ can have both positive & negative signs ∴ Δ = ± 4. Putting x1 = −2, y1 = 0, x2 = 0, y2 = 4, x3 = 0 y3 = k ± 4 = 12 −2010410k1 ± 4 = 12 −2 41k1−0 0101+1 040k ± 4 = 12 −2 4−𝑘−0 0−0+1(0−0) ± 4 = 12 −2(4−𝑘) ± 4 = – 1 (4 –k) ± 4 = – 4 + k So, 4 = – 4 + k and – 4 = – 4 + k

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.