Yes it can
The greatest number of intersection points with just four lines is 6.
The greatest number of intersection points that four coplanar lines can have occurs when no two lines are parallel and no three lines intersect at the same point. In this case, the maximum number of intersection points can be calculated using the formula ( \frac{n(n-1)}{2} ), where ( n ) is the number of lines. For four lines, this results in ( \frac{4(4-1)}{2} = 6 ) intersection points.
The greatest number is six.
- If you're working on a single sheet of paper (2-D), then you can draw four lines that intersect in 1, 2, 3, 4, 5, or 6 points. - If in 3-D space, then you can also draw four lines that don't intersect at all.
No. In plane geometry, two lines can intersect at at most one point. This means that for n points, the maximum number of intersections is limited by the number of pairs of lines. For lines a,b,c,d, there are six pairs: (a,b), (a,c), (a,d), (b,c), (b,d), (c,d). So four lines can have at most six intersections.
The greatest number of intersection points with just four lines is 6.
four
The greatest number is six.
No, four lines can can have 6 total intersections. They would be Ab, Ac, Ad, Bc, Bd, and Cd
discuss the possible number of points of interscetion of two distinct circle
- If you're working on a single sheet of paper (2-D), then you can draw four lines that intersect in 1, 2, 3, 4, 5, or 6 points. - If in 3-D space, then you can also draw four lines that don't intersect at all.
you have to make four or five horizontal lines in order to get 100 points.
On a five point star
6! Thank you to ChaCha Answers. Wiki sucks, compared to ChaCha. :) -Geometry Student from TX
No. In plane geometry, two lines can intersect at at most one point. This means that for n points, the maximum number of intersections is limited by the number of pairs of lines. For lines a,b,c,d, there are six pairs: (a,b), (a,c), (a,d), (b,c), (b,d), (c,d). So four lines can have at most six intersections.
They are four lines that coincide pairwise in at least four distinct points.
15 Consider one of the points. Call it point A. You can draw one line containing A through each of the other five lines (i.e., there are five lines that contain both A and another of the five points). Now, consider another of the points -- call it B. Excluiding the line that contains A and B, there are four lines that can be drawn containing B and one of the other four points. Continue this process for all the points. You get 5+4+3+2+1=15 lines. In general, if you have n non-collinear points, there are n+(n-1)+(n-2)+...+2+1=n*(n+1)/2 lines that can be drawn through any two of those points.