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.