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