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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 4.2, 2 (Introduction) Show that points A (a , b + c), B (b,c + a), C (c,a + b) are collinear 3 points collinear Area of triangle = 0 Area of triangle ≠ 0 Ex 4.2, 2 Show that points A (a , b + c), B (b, c + a), C (c,a + b) are collinear Three point are collinear if they lie on some line 𝑖.𝑒. They do not form a triangle ∴ Area of triangle = 0 We know that Area of triangle is given by ∆ = 1/2 |■8(x1&y1&1@x2&y2&1@x3&y3&1)| Here, x1 = a, y1 = b + c, x2 = b, y2 = c + a, x3 = c , y3 = a + b Putting values ∆ = 1/2 |■8(a&b+c&1@b&c+a&1@c&a+b&1)| Applying C1 → C1 + C2 ∆ = 1/2 |■8(a+b+c&b+c&1@b+c+a&c+a&1@c+a+b&a+b&1)| Taking (a + b + c) common from C1 ∆ = 1/2 (a + b + c) |■8(1&b+c&1@1&c+a&1@1&a+b&1)| Here, 1st and 3rd Column are Identical Hence value of determinant is zero ∆ = 1/2 (a + b + c) × 0 ∆ = 0 So, ∆ = 0 Hence points A, B & C are collinear

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.