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the distance between the two lines can be found by locating two points (one on each line) that lie on a common perpendicular to the parallel lines and calculating the distance between them. Since the lines have slope m , a common perpendicular would have slope −1/ m and we can take the line with equation y = − x / m as a common perpendicular.
The converse of the theorem implies that a homothety transforms a line in a parallel line. Conversely, the direct statement of the intercept theorem implies that a geometric transformation is always a homothety of center O, if it fixes the lines passing through O and transforms every other line into a parallel line.
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 number of vertices is smaller when some lines are parallel, or when some vertices are crossed by more than two lines. [4] An arrangement can be rotated, if necessary, to avoid axis-parallel lines. After this step, each ray that forms an edge of the arrangement extends either upward or downward from its endpoint; it cannot be horizontal.
This occurs if the lines are parallel, or if they intersect each other. Two lines that are not coplanar are called skew lines . Distance geometry provides a solution technique for the problem of determining whether a set of points is coplanar, knowing only the distances between them.
In the ordinary Euclidean plane, two lines typically intersect at a single point, but there are some pairs of lines (namely, parallel lines) that do not intersect. A projective plane can be thought of as an ordinary plane equipped with additional "points at infinity" where parallel lines intersect.
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 line with equation ax + by + c = 0 has slope -a/b, so any line perpendicular to it will have slope b/a (the negative reciprocal). Let (m, n) be the point of intersection of the line ax + by + c = 0 and the line perpendicular to it which passes through the point (x 0, y 0). The line through these two points is perpendicular to the original ...