There are m points on the line AB and n points on the line AC, excluding the point A. Triangles are formed joining these points

Options

A. `(m+n+2)/(2)`
B. `(m+n-2)/(2)`
C. `(m+n-2)/(m+n)`
D. `(m+n+2)/(m+n)`

Answer

To solve the problem of forming triangles with points on lines AB and AC, we can break down the solution step by step.

### Step 1: Understand the Problem
We have:
- \( m \) points on line AB (excluding point A).
- \( n \) points on line AC (excluding point A).

We need to form triangles using these points.

### Step 2: Identify Cases for Triangle Formation
Triangles can be formed in two distinct ways:
1. **Case 1**: Choose 2 points from line AB and 1 point from line AC.
2. **Case 2**: Choose 2 points from line AC and 1 point from line AB.

### Step 3: Calculate Combinations for Each Case
For each case, we will use combinations to determine the number of ways to select points.

**Case 1**:
- The number of ways to choose 2 points from \( m \) points on line AB is given by \( \binom{m}{2} \).
- The number of ways to choose 1 point from \( n \) points on line AC is given by \( \binom{n}{1} \).

Thus, the total combinations for Case 1 is:
\[ \text{Total for Case 1} = \binom{m}{2} \times \binom{n}{1} \]

**Case 2**:
- The number of ways to choose 2 points from \( n \) points on line AC is given by \( \binom{n}{2} \).
- The number of ways to choose 1 point from \( m \) points on line AB is given by \( \binom{m}{1} \).

Thus, the total combinations for Case 2 is:
\[ \text{Total for Case 2} = \binom{n}{2} \times \binom{m}{1} \]

### Step 4: Write the Total Combinations
The total number of triangles that can be formed is the sum of the combinations from both cases:
\[ \text{Total Triangles} = \binom{m}{2} \times \binom{n}{1} + \binom{n}{2} \times \binom{m}{1} \]

### Step 5: Substitute the Combinations
Now, substituting the values of combinations:
- \( \binom{m}{2} = \frac{m(m-1)}{2} \)
- \( \binom{n}{2} = \frac{n(n-1)}{2} \)
- \( \binom{m}{1} = m \)
- \( \binom{n}{1} = n \)

Thus, we have:
\[ \text{Total Triangles} = \left(\frac{m(m-1)}{2} \times n\right) + \left(\frac{n(n-1)}{2} \times m\right) \]

### Step 6: Simplify the Expression
Combining the two terms:
\[ \text{Total Triangles} = \frac{m(m-1)n}{2} + \frac{n(n-1)m}{2} \]
Factoring out \( \frac{mn}{2} \):
\[ \text{Total Triangles} = \frac{mn}{2} \left((m-1) + (n-1)\right) \]
\[ = \frac{mn}{2} (m + n - 2) \]

### Final Answer
Thus, the total number of triangles that can be formed is:
\[ \text{Total Triangles} = \frac{mn}{2} (m + n - 2) \]

---