The area of the triangle with vertices at the point `(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 its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step 1: Identify the coordinates
Let:
- \( (x_1, y_1) = (a, b+c) \)
- \( (x_2, y_2) = (b, c+a) \)
- \( (x_3, y_3) = (c, a+b) \)

### Step 2: Substitute the coordinates into the area formula
Substituting the coordinates into the area formula, we get:

\[ \text{Area} = \frac{1}{2} \left| a((c+a) - (a+b)) + b((a+b) - (b+c)) + c((b+c) - (c+a)) \right| \]

### Step 3: Simplify the expressions
Now, simplify each term inside the absolute value:

1. For the first term:
\[ a((c+a) - (a+b)) = a(c - b) \]

2. For the second term:
\[ b((a+b) - (b+c)) = b(a - c) \]

3. For the third term:
\[ c((b+c) - (c+a)) = c(b - a) \]

Putting it all together, we have:

\[ \text{Area} = \frac{1}{2} \left| a(c-b) + b(a-c) + c(b-a) \right| \]

### Step 4: Combine the terms
Now, combine the terms:

\[ = \frac{1}{2} \left| ac - ab + ba - bc + cb - ca \right| \]

### Step 5: Rearrange and simplify
Notice that \( ac \) and \( -ca \) cancel each other out, \( ab \) and \( -ab \) cancel each other out, and \( bc \) and \( -bc \) cancel each other out:

\[ = \frac{1}{2} \left| 0 \right| = 0 \]

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

### Final Answer
The area of the triangle is \( 0 \).

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