To find the total number of points of intersection, we use combinatorics principles by analyzing the groups of lines.
There are a total of 10 distinct lines, which can be divided into two groups:
Parallel Lines: 5 lines (L2,L4,L6,L8,L10)
Concurrent Lines: 5 lines (L1,L3,L5,L7,L9) passing through a common point C.
Step 1: Calculate maximum possible intersections
If all 10 lines were completely random and non-parallel, selecting any 2 lines would yield exactly 1 unique point of intersection:
Total possible intersections=(210)=210×9=45
Step 2: Subtract intersections lost due to parallel lines
Since the 5 parallel lines never intersect each other, we must subtract the number of points they would have normally formed:
Intersections lost (parallel)=(25)=25×4=10
Step 3: Subtract intersections lost due to concurrent lines
The 5 concurrent lines pass through a single common point C. Normally, 5 lines would intersect at multiple distinct points, but here they collapse into just 1 single point.
First, subtract all theoretical intersections among these 5 lines:
Intersections lost (concurrent)=(25)=25×4=10
Then, add back the 1 actual common point of intersection (C) where they meet:
Actual concurrent point=1
Step 4: Calculate the final number of unique intersections
Now, combine all the values together:
Number of points of intersection=(210)−(25)−(25)+1
Number of points of intersection=45−10−10+1
Number of points of intersection=25+1=26
Correct Answer: C) 26
