Permutations & Combinations
Counting intersection points of lines
Grade 11
Question:
<p>If <i>n</i> lines are drawn in a plane such that no two of them are parallel and no three of them are concurrent, such that these lines divide the plane into 67 parts, then find the number of different points at which these lines will cut.</p>
Step-by-Step Solution
Key Concept: Use the formula for regions created by n lines in general position, then find intersection points using combinations.
<p><strong>Step 1:</strong> When <i>n</i> lines are drawn in a plane with no two parallel and no three concurrent, the number of parts into which they divide the plane is given by:</p><p>$$1 + \frac{n(n^2 + n + 2)}{2}$$</p><p><strong>Step 2:</strong> Setting this equal to 67:</p><p>$$1 + \frac{n(n^2 + n + 2)}{2} = 67$$</p><p>$$\frac{n(n^2 + n + 2)}{2} = 66$$</p><p>$$n^2 + n - 132 = 0$$</p><p><strong>Step 3:</strong> Factoring:</p><p>$$(n + 12)(n - 11) = 0$$</p><p>$$n = 11 \text{ (since } n > 0\text{)}$$</p><p><strong>Step 4:</strong> The number of intersection points when no two lines are parallel and no three are concurrent is:</p><p>$$\binom{n}{2} = \binom{11}{2} = \frac{11 \times 10}{2} = 55$$</p><p>∴ The answer is <strong>55</strong>.</p>
Correct Answer: 55