# Show that pointsA (a, b + c), B (b, c + a), C (c, a + b) are collinear.

Given vertices of the triangle are A (a, b + c), B (b, c + a), C (c, a + b)

Let the vertices of the triangle be given by (x1, y1), (x2, y2), (x3, y3)

Area of triangle is given by Δ = For points to be collinear area of triangle = Δ = 0

So, we have to show that area of triangle formed by ABC is 0

Area of triangle = Δ = Expanding the determinant along Row 1

Δ = 1/2 × [a × {(c + a) × 1 – (a + b) × 1) – (b + c) × {b × 1 – c × 1} + 1 × {b × (a + b) – c × (c +a)}]

Δ = 1/2 × [a × (c + a – a – b) – (b + c) × (b – c) + 1 × (ab + b2 – c2 - ca)]

Δ = 1/2 × [a × (c – b) – (b2 – c2) + 1 × (ab + b2 – c2 – ca)]

Δ = 1/2 × (ac – ab – b2 + c2 + ab + b2 – c2 – ca) sq units

Δ = 1/2 × 0 sq units

Δ = 0

Given vertices of the triangle are A (a, b + c), B (b, c + a), C (c, a + b) are collinear

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