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How many intersects are there for 4 lines? - Answers

Maximum 12 intersections are there! You can simply use the formula: No. of intersections = n(n-1)/2 [where 'n' is the number of lines] This is derived as each new line can intersect (at most) all the previously drawn lines. There if there is: 1 line => 0 intersections 2 lines => 1 intersection 3 lines => 3 intersections [1 that was there + 2 by the new line can intersect both the previous lines.] 4 lines => 6 intersections [3 that were already there + 3 because the new line can intersect all the 3 lines that were present previously.]



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How many intersects are there for 4 lines? - Answers

https://math.answers.com/math-and-arithmetic/How_many_intersects_are_there_for_4_lines

Maximum 12 intersections are there! You can simply use the formula: No. of intersections = n(n-1)/2 [where 'n' is the number of lines] This is derived as each new line can intersect (at most) all the previously drawn lines. There if there is: 1 line => 0 intersections 2 lines => 1 intersection 3 lines => 3 intersections [1 that was there + 2 by the new line can intersect both the previous lines.] 4 lines => 6 intersections [3 that were already there + 3 because the new line can intersect all the 3 lines that were present previously.]



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https://math.answers.com/math-and-arithmetic/How_many_intersects_are_there_for_4_lines

How many intersects are there for 4 lines? - Answers

Maximum 12 intersections are there! You can simply use the formula: No. of intersections = n(n-1)/2 [where 'n' is the number of lines] This is derived as each new line can intersect (at most) all the previously drawn lines. There if there is: 1 line => 0 intersections 2 lines => 1 intersection 3 lines => 3 intersections [1 that was there + 2 by the new line can intersect both the previous lines.] 4 lines => 6 intersections [3 that were already there + 3 because the new line can intersect all the 3 lines that were present previously.]

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      Maximum 12 intersections are there! You can simply use the formula: No. of intersections = n(n-1)/2 [where 'n' is the number of lines] This is derived as each new line can intersect (at most) all the previously drawn lines. There if there is: 1 line => 0 intersections 2 lines => 1 intersection 3 lines => 3 intersections [1 that was there + 2 by the new line can intersect both the previous lines.] 4 lines => 6 intersections [3 that were already there + 3 because the new line can intersect all the 3 lines that were present previously.]

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