Using determinants show that points $A (a, b + c), B (b, c + a)$ and $C (c, a + b)$ are col-linear.

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Solution

For the points to be collinear the area of the triangle formed by the points must be zero.
Therefore,
$∣ ∣ ∣ ∣ \begin{matrix} a & b + c & 1 b & c + a & 1 c & a + b & 1 \end{matrix} ∣ ∣ ∣ ∣$ must be equal to $0$ .
L.H.S. = $(a + b + c) ∣ ∣ ∣ ∣ \begin{matrix} 1 & b + c & 1 1 & c + a & 1 1 & a + b & 1 \end{matrix} ∣ ∣ ∣ ∣$ = $0$ . (Since, elements of two columns are equal.)
Hence, the given points are collinear.

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Similar questions

A (a, b + c), B (b, c + a)

C (c, a + b)

Using the property of determinants and without expanding, prove that:

A (a, b + c), B (b, c + a), C (c, a + b)

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a & a^{2} 1 & b & b^{2} 1 & c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (a - b) (b - c) (c - a)

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{3} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (a - b) (b - c) (c - a) (a + b + c)

A (a, b + c), B (b, c + a)

C (c, a + b)

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