In a plane there are two families of lines y=x+r and y=-x+r where r belongs to 0,1,2.3,4 intersecting to yield 25 points 
Find the number of quadrilaterals that can be formed by these points

To solve the problem, let's break it down step by step.

### 1. **Understanding the Problem:**
   We have two families of lines:
   - \( y = x + r \)
   - \( y = -x + r \)
   
   where \( r \) takes values \( 0, 1, 2, 3, 4 \). Each family has 5 lines corresponding to these 5 values of \( r \).

### 2. **Finding the Intersection Points:**
   An intersection point occurs when a line from the first family intersects with a line from the second family. To find the intersection, set:
   \[
   y = x + r_1 \quad \text{(from the first family)}
   \]
   equal to
   \[
   y = -x + r_2 \quad \text{(from the second family)}.
   \]
   
   This gives:
   \[
   x + r_1 = -x + r_2
   \]
   \[
   2x = r_2 - r_1
   \]
   \[
   x = \frac{r_2 - r_1}{2}
   \]
   
   Substituting \( x \) back into either equation (say \( y = x + r_1 \)), we get:
   \[
   y = \frac{r_2 - r_1}{2} + r_1 = \frac{r_2 + r_1}{2}
   \]
   
   The point of intersection is then:
   \[
   \left(\frac{r_2 - r_1}{2}, \frac{r_2 + r_1}{2}\right)
   \]
   
   Since both \( r_1 \) and \( r_2 \) can independently take values 0, 1, 2, 3, 4, there are \( 5 \times 5 = 25 \) such intersection points.

### 3. **Counting the Number of Quadrilaterals:**
   A quadrilateral can be formed by selecting 4 points out of the 25 intersection points. The number of ways to choose 4 points out of 25 is given by the combination formula:
   \[
   \binom{25}{4} = \frac{25 \times 24 \times 23 \times 22}{4 \times 3 \times 2 \times 1} = 12650
   \]
   
   Thus, the number of quadrilaterals that can be formed is **12,650**.

### Conclusion:
The total number of quadrilaterals that can be formed by the intersection points is **12,650**.

---

Would you like a detailed explanation of any step or have any other questions? Here are some related problems you might find interesting:

1. How many triangles can be formed by these 25 points?
2. What if \( r \) took different values, say \( r \) belonged to a different set?
3. How does the problem change if only three lines from each family were considered?
4. What if we had three families of lines, say \( y = ax + r \) where \( a \) can also vary?
5. Can we form any special quadrilaterals (like squares or rectangles) from these points?

**Tip:** When working with combinatorial problems, it’s helpful to break them down into smaller, manageable parts.

In a plane there are two families of lines y=x+r and y=-x+r where r belongs to 0,1,2,3,4 intersecting to yield 25 points. Find the number of quadrilaterals that can be formed by these points.

Explore how to calculate the number of quadrilaterals formed by 25 intersection points of two families of lines y=x+r and y=-x+r, where r belongs to {0,1,2,3,4}. This problem uses combinatorics and geometry, suitable for high school students in Grades 9-12.

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