There are m points on a straight line AB & n points on the line AC none of them being the point A. Tri- angles are formed with these points as vertices, when
A is excluded (ii) A is included.
The ratio of number of triangles in the two cases is

Options

A. `(m + n - 2)/(m + n)`
B. `(m + n - 2)/(m + n - 1)`
C. `(m + n -2)/(m + n +2)`
D. `(m(n - 1))/((m + 1)(n + 1))`

Answer

To solve the problem, we need to find the number of triangles formed by points on lines AB and AC, both when point A is excluded and when it is included. Let's break this down step by step.

### Step 1: Count triangles when A is excluded

When point A is excluded, we can only use the m points on line AB and the n points on line AC.

1. **Case 1**: Select 2 points from line AB and 1 point from line AC.
- The number of ways to select 2 points from m points on line AB is given by the combination formula \( \binom{m}{2} \).
- The number of ways to select 1 point from n points on line AC is \( \binom{n}{1} \).
- Therefore, the number of triangles formed in this case is:
\[ T_1 = \binom{m}{2} \cdot \binom{n}{1} = \frac{m(m-1)}{2} \cdot n \]

2. **Case 2**: Select 2 points from line AC and 1 point from line AB.
- The number of ways to select 2 points from n points on line AC is \( \binom{n}{2} \).
- The number of ways to select 1 point from m points on line AB is \( \binom{m}{1} \).
- Therefore, the number of triangles formed in this case is:
\[ T_2 = \binom{n}{2} \cdot \binom{m}{1} = \frac{n(n-1)}{2} \cdot m \]

3. **Total triangles when A is excluded**:
\[ T_{exclude} = T_1 + T_2 = \frac{m(m-1)}{2} \cdot n + \frac{n(n-1)}{2} \cdot m \]

### Step 2: Count triangles when A is included

When point A is included, we can choose to either include A in the triangle or not.

1. **Case 1**: Include point A and select 1 point from line AB and 1 point from line AC.
- The number of ways to select 1 point from line AB is \( \binom{m}{1} \).
- The number of ways to select 1 point from line AC is \( \binom{n}{1} \).
- Therefore, the number of triangles formed in this case is:
\[ T_{A} = \binom{m}{1} \cdot \binom{n}{1} = m \cdot n \]

2. **Case 2**: Exclude A and use the triangles counted in the previous step.
- This is simply \( T_{exclude} \).

3. **Total triangles when A is included**:
\[ T_{include} = T_{A} + T_{exclude} = m \cdot n + \left( \frac{m(m-1)}{2} \cdot n + \frac{n(n-1)}{2} \cdot m \right) \]

### Step 3: Calculate the ratio of triangles

Now we need to find the ratio of the number of triangles when A is excluded to the number of triangles when A is included:
\[ \text{Ratio} = \frac{T_{exclude}}{T_{include}} \]

Substituting the values we found:
\[ \text{Ratio} = \frac{\frac{m(m-1)}{2} \cdot n + \frac{n(n-1)}{2} \cdot m}{m \cdot n + \left( \frac{m(m-1)}{2} \cdot n + \frac{n(n-1)}{2} \cdot m \right)} \]

### Step 4: Simplify the ratio

Let's simplify this expression:
1. The numerator is \( \frac{m(m-1)}{2} \cdot n + \frac{n(n-1)}{2} \cdot m \).
2. The denominator is \( m \cdot n + \frac{m(m-1)}{2} \cdot n + \frac{n(n-1)}{2} \cdot m \).

After simplification, we find:
\[ \text{Ratio} = \frac{m + n - 2}{m + n} \]

### Final Answer

Thus, the ratio of the number of triangles formed when A is excluded to the number of triangles formed when A is included is:
\[ \frac{m + n - 2}{m + n} \]