<p>There are <em>n</em> straight lines in a plane, in which no two are parallel and no three pass through the same point. Their points of intersection are joined. Show that the number of new lines thus introduced is</p><p>\[\frac{1}{8}n(n-1)(n-2)(n-3)\]</p>
Step-by-Step Solution
Key Concept: A new line is formed by joining two intersection points, which are themselves formed by two different pairs of original lines. The key is recognizing that each new line requires selecting 4 distinct original lines (2 lines for each intersection point), and we must avoid collinearity by ensuring the 4 lines form a proper configuration.
<p><strong>Step 1:</strong> Total intersection points formed by n lines with no two parallel and no three concurrent = C(n,2) = n(n-1)/2</p><p><strong>Step 2:</strong> Naively, joining any two intersection points could give C(n(n-1)/2, 2) potential lines. However, many pairs of intersection points are collinear on the original n lines.</p><p><strong>Step 3:</strong> Count collinear triples: Each original line passes through C(n-1, 2) intersection points (formed by that line with the other n-1 lines). So total collinear pairs on original lines = n·C(n-1, 2) = n(n-1)(n-2)/2</p><p><strong>Step 4:</strong> Each such collinear pair on an original line does NOT form a new line. Number of new lines = C(n(n-1)/2, 2) - n·C(n-1, 2)</p><p><strong>Step 5:</strong> Calculate: C(n(n-1)/2, 2) = [n(n-1)/2]·[n(n-1)/2 - 1]/2 = n(n-1)[n(n-1)-2]/8</p><p><strong>Step 6:</strong> Subtract the collinear pairs: n(n-1)[n(n-1)-2]/8 - n(n-1)(n-2)/2 = n(n-1)/8·[(n(n-1)-2) - 4(n-2)] = n(n-1)/8·[n² - n - 2 - 4n + 8] = n(n-1)(n² - 5n + 6)/8 = n(n-1)(n-2)(n-3)/8</p><p>∴ Answer: <strong>1/8·n(n-1)(n-2)(n-3)</strong></p>
Correct Answer: \frac{1}{8}n(n-1)(n-2)(n-3)