Q 08. Two lines AB and AC join at point A. If total number of points on line AB and AC are 7 and 9, then find the number of distinct triangle that can be formed using these points.

Ops: A. 226

B. 336

C. 352

D. 402

To solve this problem, we need to find the number of distinct triangles that can be formed using the points on the two lines \( AB \) and \( AC \), which share the common point \( A \).

### Step-by-step Approach:

1. **Understanding the Problem:**
   - There are 7 points on line \( AB \), including point \( A \).
   - There are 9 points on line \( AC \), including point \( A \).
   - A triangle is formed by selecting 3 distinct points, and those points should not all lie on the same straight line (collinear).

2. **Counting the Total Points:**
   - The total number of points is:
     \[
     7 \ (\text{points on AB}) + 9 \ (\text{points on AC}) - 1 \ (\text{subtract A, as it is double-counted}) = 15 \ \text{points}.
     \]
   - These points include point \( A \), which is shared by both lines.

3. **Calculating Total Triangles:**
   - The total number of triangles that can be formed by selecting any 3 points from the 15 available points is:
     \[
     \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455.
     \]

4. **Subtracting Collinear Points:**
   - We need to subtract the number of invalid triangles, i.e., the triangles formed by points that all lie on the same line (collinear).
   - On line \( AB \), the number of ways to choose 3 points (which would be collinear) is:
     \[
     \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35.
     \]
   - On line \( AC \), the number of ways to choose 3 collinear points is:
     \[
     \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84.
     \]
   - So, the total number of collinear triangles is:
     \[
     35 + 84 = 119.
     \]

5. **Finding the Number of Valid Triangles:**
   - The number of distinct triangles is the total number of triangles minus the collinear triangles:
     \[
     455 - 119 = 336.
     \]

Thus, the number of distinct triangles that can be formed is **336**.

### Final Answer:  
**Option B: 336**

Would you like more details or have any questions about this solution? Here are some related questions for deeper understanding:

1. How do you calculate combinations using \(\binom{n}{k}\)?
2. What is the significance of subtracting the collinear triangles?
3. How would the answer change if a different number of points were given on each line?
4. Can the triangle include point \( A \)? Why or why not?
5. What if the points formed a polygon instead of two lines? How would the triangle counting change?

**Tip:** When counting combinations, always check for constraints like collinearity that might reduce the number of valid solutions!

Two lines AB and AC join at point A. If total number of points on line AB and AC are 7 and 9, then find the number of distinct triangles that can be formed using these points. A. 226 B. 336 C. 352 D. 402

Learn how to find the number of distinct triangles that can be formed using points from two lines AB and AC. This problem involves combinatorics and geometry, with a focus on counting valid triangles by eliminating collinear points. Suitable for high school students in Grades 10-12.

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