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Let l 1 = [a 1, b 1, c 1] and l 2 = [a 2, b 2, c 2] be a pair of distinct lines. Then the intersection of lines l 1 and l 2 is point a P = (x 0, y 0, z 0) that is the simultaneous solution (up to a scalar factor) of the system of linear equations: a 1 x + b 1 y + c 1 z = 0 and a 2 x + b 2 y + c 2 z = 0. The solution of this system gives: x 0 ...
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]
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 ...
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 ...
Each such pair has a unique intersection point in the extended Euclidean plane. 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.
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 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 .
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