Math Problem Statement

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

Solution

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

1. Understanding the Problem:

We have two families of lines:

  • y=x+ry = x + r
  • y=x+ry = -x + r

where rr takes values 0,1,2,3,40, 1, 2, 3, 4. Each family has 5 lines corresponding to these 5 values of rr.

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+r1(from the first family)y = x + r_1 \quad \text{(from the first family)} equal to y=x+r2(from the second family).y = -x + r_2 \quad \text{(from the second family)}.

This gives: x+r1=x+r2x + r_1 = -x + r_2 2x=r2r12x = r_2 - r_1 x=r2r12x = \frac{r_2 - r_1}{2}

Substituting xx back into either equation (say y=x+r1y = x + r_1), we get: y=r2r12+r1=r2+r12y = \frac{r_2 - r_1}{2} + r_1 = \frac{r_2 + r_1}{2}

The point of intersection is then: (r2r12,r2+r12)\left(\frac{r_2 - r_1}{2}, \frac{r_2 + r_1}{2}\right)

Since both r1r_1 and r2r_2 can independently take values 0, 1, 2, 3, 4, there are 5×5=255 \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: (254)=25×24×23×224×3×2×1=12650\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 rr took different values, say rr 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+ry = ax + r where aa 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.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Geometry
Linear Equations
Intersection Points

Formulas

Intersection point formula: x = (r2 - r1)/2, y = (r2 + r1)/2
Combination formula: C(n, k) = n! / (k!(n-k)!)

Theorems

Combination Theorem
Geometric Interpretation of Line Intersections

Suitable Grade Level

Grades 9-12