Given:
- Lines \( L_{2n-1} \) (\( n = 1, 2, \dots, 10 \)) are parallel to each other.
- Lines \( L_{2n} \) (\( n = 1, 2, \dots, 10 \)) pass through a common point \( P \).
Step 1: Points of Intersection between \( L_{2n-1} \) and \( L_{2m} \)
Since all \( L_{2n-1} \) lines are parallel, they do not intersect among themselves. Similarly, all \( L_{2n} \) lines pass through the point \( P \), so they intersect at \( P \) and do not form additional intersection points among themselves.
However, each line \( L_{2n-1} \) intersects each line \( L_{2m} \) exactly once (since they are not parallel), leading to:
\[ 10 \times 10 = 100 \text{ intersection points} \]
Step 2: Points of Intersection among \( L_{2n} \) Lines
All \( L_{2n} \) lines pass through the common point \( P \). Therefore, there is exactly one intersection point among these lines at \( P \).
Step 3: Total Number of Points of Intersection
The total number of points of intersection is given by:
\[ 100 + 1 = 101 \]
Conclusion: The maximum number of points of intersection of pairs of lines from the set \( \{L_1, L_2, \dots, L_{20}\} \) is 101.
Match List-I with List-II
List-I | List-II |
---|---|
(A) \(^{8}P_{3} - ^{10}C_{3}\) | (I) 6 |
(B) \(^{8}P_{5}\) | (II) 21 |
(C) \(^{n}P_{4} = 360,\) then find \(n\). | (III) 216 |
(D) \(^{n}C_{2} = 210,\) find \(n\). | (IV) 6720 |
Choose the correct answer from the options given below:
The net current flowing in the given circuit is ___ A.
If the equation \( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) has equal roots, where \( a + c = 15 \) and \( b = \frac{36}{5} \), then \( a^2 + c^2 \) is equal to .