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The intersection point falls within the first line segment if 0 ≤ t ≤ 1, and it falls within the second line segment if 0 ≤ u ≤ 1. These inequalities can be tested without the need for division, allowing rapid determination of the existence of any line segment intersection before calculating its exact point. [3]
The intersection (red) of two disks (white and red with black boundaries). The circle (black) intersects the line (purple) in two points (red). The disk (yellow) intersects the line in the line segment between the two red points. The intersection of D and E is shown in grayish purple. The intersection of A with any of B, C, D, or E is the empty ...
If one wants to determine the intersection points of two polygons, one can check the intersection of any pair of line segments of the polygons (see above). For polygons with many segments this method is rather time-consuming. In practice one accelerates the intersection algorithm by using window tests. In this case one divides the polygons into ...
Lines that meet at the same point are said to be concurrent. The set of all lines in a plane incident with the same point is called a pencil of lines centered at that point. The computation of the intersection of two lines shows that the entire pencil of lines centered at a point is determined by any two of the lines that intersect at that point.
For each pair of lines, there can be only one cell where the two lines meet at the bottom vertex, so the number of downward-bounded cells is at most the number of pairs of lines, () /. Adding the unbounded and bounded cells, the total number of cells in an arrangement can be at most n ( n + 1 ) / 2 + 1 {\displaystyle n(n+1)/2+1} . [ 5 ]
the distance between the two lines is the distance between the two intersection points of these lines with the perpendicular line y = − x / m . {\displaystyle y=-x/m\,.} This distance can be found by first solving the linear systems
The de Longchamps point is the point of concurrence of several lines with the Euler line. Three lines, each formed by drawing an external equilateral triangle on one of the sides of a given triangle and connecting the new vertex to the original triangle's opposite vertex, are concurrent at a point called the first isogonal center .
Monge's theorem states that the three such points given by the three pairs of circles always lie in a straight line. In the case of two of the circles being of equal size, the two external tangent lines are parallel. In this case Monge's theorem asserts that the other two intersection points must lie on a line parallel to those two external ...