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Ex 4.3, 2 - Show that A (a , b + c), B (b,c + a), C (c,a + b) - Area of triangle

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Ex 4.3, 2 (Introduction) Show that points A (a , b + c), B (b,c + a), C (c,a + b) are collinear Ex 4.3, 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 ⇒ They do not form a triangle ⇒ Area of triangle = 0 We know that Area of triangle is given by ∆ = 1﷮2﷯ 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 ∆ = 1﷮2﷯ a﷮b+c﷮1﷮b﷮c+a﷮1﷮c﷮a+b﷮1﷯﷯ Applying C1 →C1 + C2 ∆ = 1﷮2﷯ 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) 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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