Question

Total number of regions in which 'n' coplanar lines can divide the plane, it is known that no two lines are parallel and no three of them are concurrent, is equal to

• A. \(\frac{1}{2}(n^2 + n + 2)\)
• B. \(\frac{1}{2}(n + 3n^2)\)
• C. \(\frac{1}{2}(3n + n^2)\)
• D. \((n^2 - n + 2)\)

Hint:

The maximum number of regions from n lines is \(\frac{n(n+1)}{2} + 1\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The maximum number of regions formed by \(n\) lines is given by a recurrence relation.

Step 2: Detailed Explanation:
Let \(R(n)\) be the number of regions. When the nth line is added, it intersects the previous \(n-1\) lines at \(n-1\) distinct points, dividing the line into \(n\) segments. Each segment divides an existing region into two, adding \(n\) new regions. So \(R(n) = R(n-1) + n\) with \(R(0) = 1\). Then \(R(n) = 1 + \sum_{k=1}^n k = 1 + \frac{n(n+1)}{2} = \frac{n^2 + n + 2}{2}\).

Step 3: Final Answer:
\(\frac{1}{2}(n^2 + n + 2)\), which corresponds to option (A).