For real numbers a, b and c, the area of the triangle with vertices \((a, b + c)\), \((b, c + a)\) and \((c, a + b)\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
\(\displaystyle \Delta = \frac{1}{2}\begin{vmatrix}a & b + c & 1\\ b & c + a & 1\\ c & a + b & 1\end{vmatrix}\)
Under the column operation \(C_1 \to C_1 + C_2\), every entry of the first column becomes \(a + b + c\):
\(\displaystyle \Delta = \frac{1}{2}\begin{vmatrix}a + b + c & b + c & 1\\ a + b + c & c + a & 1\\ a + b + c & a + b & 1\end{vmatrix}\)
\(\displaystyle \Delta = \frac{1}{2}(a + b + c)\begin{vmatrix}1 & b + c & 1\\ 1 & c + a & 1\\ 1 & a + b & 1\end{vmatrix}\)
The first and third columns are now identical, so the determinant is 0.
\(\Delta = 0\)
The three points always lie on the line x + y = a + b + c, so they are collinear and the area is 0.