Permutations & Combinations
Counting Intersections
Grade 11
Question:
<p>Consider 5 points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or co incident. From each point perpendiculars are drawn to all the lines joining the other four points. If \(k\) is the maximum number of intersections that these perpendiculars can have? Then</p>
<p>more than 310</p>
<p>\(k\) is divisible by 5</p>
<p>\(k\) is less than 400</p>
<p>\(k\) is even number</p>
Step-by-Step Solution
Key Concept: Each of the 5 points generates perpendiculars to C(4,2)=6 lines, giving 5×6=30 perpendiculars total. The maximum intersections occur when no three perpendiculars are concurrent and no two are parallel (guaranteed by the given conditions), so we count all pairwise intersections: C(30,2).
<p><strong>Step 1:</strong> Count total perpendiculars drawn.</p><p>From each of 5 points, perpendiculars are drawn to lines joining the other 4 points. The number of such lines from any point = C(4,2) = 6.</p><p>Total perpendiculars = 5 × 6 = 30.</p><p><strong>Step 2:</strong> Determine maximum intersections.</p><p>Given conditions (no two lines parallel, perpendicular, or coincident) ensure:</p><ul><li>No two perpendiculars are parallel</li><li>No three perpendiculars meet at a point (general position)</li></ul><p><strong>Step 3:</strong> Count all pairwise intersections.</p><p>Maximum intersections = C(30,2) = (30 × 29)/2 = 435.</p><p>∴ Answer: k = 435 (Option A)</p>
Correct Answer: A