Question

Show that points A (a,b+c),B (b,c+a),C (c,a+b) are collinear

Answer 1

Area of ∆ ABC is given by the relation,

△=\frac{1}{2}$$\begin{vmatrix}a&b+c&1\\\b&c+a&1\\\c&a+b&1\end{vmatrix}

=\frac{1}{2}$$\begin{vmatrix}a&b+c&1\\\b-a&a-b&0\\\c-a&a-c&0\end{vmatrix} (Applying R2$\to$ R2-R1 and R33$\to$R3-R1)

=$\frac{1}{2}$(a-b)(c-a)$\begin{vmatrix}a&b+c&1\\\\-1&1&0\\\1&-1&0\end{vmatrix}$

=$\frac{1}{2}$(a-b)(c-a)$\begin{vmatrix}a&b+c&1\\\\-1&1&0\\\1&-1&0\end{vmatrix}$ (applying R3$\to$ R3+R2)
=0 (All elements of R3 are 0)
Thus, the area of the triangle formed by points A, B, and C is zero

Hence, the points A, B, and C are collinear.