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Question
Write the area of the triangle with vertices at (a, b + c), (b, c + a) and (c, a + b).
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Solution
Let A(a, b + c), B(b, c + a) and C(c, a + b) be the vertices of the the given triangle.
\[\therefore\text{ Area of ∆ ABC }= \frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right|\]
\[ = \frac{1}{2}\left| a\left( c + a - a - b \right) + b\left( a + b - b - c \right) + c\left( b + c - c - a \right) \right|\]
\[ = \frac{1}{2}\left| a\left( c - b \right) + b\left( a - c \right) + c\left( b - a \right) \right|\]
\[ = \frac{1}{2}\left| ac - ab + ab - bc + bc - ac \right|\]
\[ = 0\]
Hence, area of the triangle with vertices at (a, b + c), (b, c + a) and (c, a + b) is 0.
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