Q. 114.6( 8 Votes )

# Show that points

A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

Answer :

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 (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

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 + b^{2} – c^{2} - ca)]

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

Δ = 1/2 × (ac – ab – b^{2} + c^{2} + ab + b^{2} – c^{2} – 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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