There are 8 lines in a plane, no two of which are parallel. What is the maximum number of points at which they can intersect?
28
Finding Maximum Intersection Points for Lines in a Plane
The question asks for the maximum number of points at which 8 lines in a plane can intersect, given that no two lines are parallel.
When no two lines are parallel, the maximum number of intersection points occurs when every pair of lines intersects at a distinct point. If there were parallel lines, they would not intersect, reducing the total number of intersection points. If more than two lines intersected at the same point, that would also reduce the total number of distinct intersection points.
To find the maximum number of intersection points, we need to determine how many unique pairs of lines can be formed from the given number of lines. Each pair of lines will intersect at exactly one point (since no two are parallel).
The number of ways to choose 2 items from a set of \(n\) items, where the order doesn't matter, is given by the combination formula:
\(C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}\)
In this problem, we have \(n=8\) lines, and we are choosing \(k=2\) lines at a time to form a pair that intersects. So, the maximum number of intersection points is \(C(8, 2)\).
Calculating the Maximum Intersection Points
Let's calculate \(C(8, 2)\):
\(C(8, 2) = \binom{8}{2} = \frac{8!}{2!(8-2)!}\)
\(C(8, 2) = \frac{8!}{2!6!}\)
\(C(8, 2) = \frac{8 \times 7 \times 6!}{2 \times 1 \times 6!}\)
We can cancel out the \(6!\) from the numerator and the denominator:
\(C(8, 2) = \frac{8 \times 7}{2 \times 1}\)
\(C(8, 2) = \frac{56}{2}\)
\(C(8, 2) = 28\)
So, the maximum number of intersection points for 8 lines in a plane, with no two lines parallel, is 28.
Summary of Calculation
| Number of Lines (\(n\)) | Pairs of Lines (Combination \(C(n, 2)\)) | Maximum Intersection Points |
|---|---|---|
| 8 | \(C(8, 2)\) | 28 |
This calculation shows that with 8 lines, where every pair intersects at a unique point, there will be a total of 28 distinct intersection points.
Revision Table: Line Intersection Concepts
| Concept | Description | Formula |
|---|---|---|
| Intersection Point | A point where two or more lines cross each other. | N/A |
| Parallel Lines | Lines in a plane that do not intersect. | Slopes are equal (for linear equations). |
| Maximum Intersection Points (n lines) | The greatest possible number of distinct points where \(n\) lines in a plane can intersect, assuming no two are parallel and no three are concurrent. | \(C(n, 2) = \frac{n(n-1)}{2}\) |
Additional Information: Understanding Combinations and Geometry
The problem connects concepts from geometry (lines and intersection points) and combinatorics (combinations).
- Lines in a Plane: A line is a straight path extending infinitely in both directions. In a plane, two distinct lines can either be parallel (never intersect) or intersect at exactly one point.
- Maximizing Intersections: To get the maximum number of intersection points, we need to ensure two conditions are met:
- No two lines are parallel. This guarantees that every pair of lines will intersect.
- No three or more lines intersect at the same single point. If multiple lines passed through the same point, they would only contribute one intersection point instead of multiple unique points from separate pairwise intersections. The problem implicitly assumes this scenario by asking for the maximum number of *points*.
- Combinations (\(C(n, k)\)): Combinations are used when we need to select a group of items from a larger set, and the order of selection doesn't matter. In this case, picking line 1 and then line 2 is the same as picking line 2 and then line 1; they form the same pair and create the same intersection point. The formula \(\frac{n(n-1)}{2}\) for \(C(n, 2)\) arises because there are \(n\) choices for the first line and \((n-1)\) choices for the second line, giving \(n(n-1)\) ordered pairs. Since the order doesn't matter, we divide by 2 to get the number of unordered pairs.
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