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Show that the points A(a, b + c), B(b, c + a) and C(c, a + b) are collinear. |
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Answer» Given points are A(a, b + c), B(b, c + a) and C(c, a + b). We know that area of triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is given by \(\frac{1}{2}\) [x1 (y2 − y3) + x2 (y3 − y1 ) + x3 (y1 − y2)]. ∴ Area of triangle formed by the points A, B and C = \(\frac{1}{2}\) [a((c + a) − (a + b)) + ((a + b) − (b+ c)) +c ((b + c)– (c + a))] = \(\frac{1}{2}\) [a(c – b) + b(a – c) + c(b – a)] = \(\frac{1}{2}\) [ac – ab + ba – bc + cb – ca] = \(\frac{1}{2}\) [ac – ca + ba – ab + cb – bc] = \(\frac{1}{2}\) [ac – ac + ba – ba + cb – cb] (∵ ac = ca & ba = ab & bc = cb) = \(\frac{1}{2}\) × 0 = 0. Area of triangle formed by the point A, B and C is zero which is condition that A, B & C are collinear. Hence, points A(a, b + c), B(b, c + a), and C(c, a + b) are collinear. |
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