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!