The area of the triangle with vertices at the points `(a,b+c),(b,c+a),(c,a+b)` is

Options

A. 0
B. `a+b+c`
C. `ab+bc+ca`
D. None of these

Answer

To find the area of the triangle with vertices at the points \((a, b+c)\), \((b, c+a)\), and \((c, a+b)\), we can use the formula for the area of a triangle given by the coordinates of its vertices.

### Step-by-Step Solution:

1. **Identify the vertices**:
The vertices of the triangle are given as:
- \( A(a, b+c) \)
- \( B(b, c+a) \)
- \( C(c, a+b) \)

2. **Use the area formula**:
The area \( A \) of a triangle with vertices at \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

3. **Substitute the coordinates**:
Substituting the coordinates of the vertices into the area formula:
\[ \text{Area} = \frac{1}{2} \left| a((c+a) - (a+b)) + b((a+b) - (b+c)) + c((b+c) - (c+a)) \right| \]

4. **Simplify the expression**:
Now, we simplify each term:
- The first term:
\[ a((c+a) - (a+b)) = a(c - b) \]
- The second term:
\[ b((a+b) - (b+c)) = b(a - c) \]
- The third term:
\[ c((b+c) - (c+a)) = c(b - a) \]

Combining these, we have:
\[ \text{Area} = \frac{1}{2} \left| a(c-b) + b(a-c) + c(b-a) \right| \]

5. **Factor the expression**:
Rearranging the terms:
\[ = \frac{1}{2} \left| ac - ab + ba - bc + cb - ca \right| \]
Notice that \( ac \) and \( -ca \) cancel out, and we are left with:
\[ = \frac{1}{2} \left| -ab + ba - bc \right| = \frac{1}{2} \left| 0 \right| = 0 \]

6. **Conclusion**:
Since the area is \( 0 \), this means that the points are collinear.

### Final Answer:
The area of the triangle with vertices at the points \((a, b+c)\), \((b, c+a)\), and \((c, a+b)\) is \(0\).