45 points
By definition, any lines/segments that are perpendicular to each other either do, or (in the case of non-intersecting segments) would, intersect each other at right angles. A right angle being a 90* angle. Therefore, perpendicular, intersecting line segments will form 4 90* angles.
their slopes are negative reciprocals of each other. and they make a right angle when they intersect.
Yes. Draw three line segments so that they cross at three points forming a triangle (with each side extending beyond the vertices of the triangle). Draw one circle to enclose the triangle without touching it to intersect the extended sides at a further 6 points, making 9 points of intersection so far. Draw the second circle slightly shifted (relative to the first) so that it also encloses the triangle (without touching it) creating a further 6 intersection points with the three lines and 2 with the first circle; an additional 8 intersection points making 17 in all.
540 degrees. Draw line segments to connect vertices and make trianlges. You will have three triangles. Since each triangle has 180 degrees and there are three of them (180 x 3) the answer is 540 degrees. A square or rectangle has four right angles of 90 degrees each (90 x 4) and therefore it is 360 degrees. You can make two triangle when you draw a diagonal line. Each triangle is 180 degrees and equals 360 degrees.
The answer will depend on the relative positions of the points.
Two is enough to connect each point to at least one other. Three is enough for an open net. Six to connect every point to each of the rest. The question should be more specific.
draw 3 line segments connect the endpoint what figure is formed
Given that an octogon has eight sides, then the simplest way would be to enscribe a circle. Divide it exactly in half, then divide it in half again, at right angles to your first line.You now have four quarters. Divide each quarter in half, and you will have eight segments. Connect the points of the segments and you will have an octogon.
If the points are arranged on an imaginary line with one line connecting each point to its neighbor, there will be 99 lines (or line segments).
If there are n points then the maximum number of lines possible is n*(n-1)/2 and that maximum is attained of no three points are collinear.
bridge
The two points on exactly opposite sides of a circle are parallel to each other. This can be evidenced by finding the derivative/gradient at those points; if they are the same then the two line segments described by those points are parallel.
Yes, it is possible to connect the 3 access point with each other. This is because of the availability of the repeating and bridging options in the three access points.
Every line and every line segment of >0 length has an inifinite amount of unique points.Socratic Explaination:consider ...- There are 2 distinct points defining a line segment.- Between these 2 distinct points, there is a midpoint.- The midpoint divides the original segment into 2 segments of equal length.- There are 2 distinct points used to define each segment.- Between these 2 distinct points, there is a midpoint for each segment.- These midpoints divide the segments into smaller segments of equal length.- repeat until throughly beatenThis leads to a description of an infinite amount of points for any given line segment.This does not describe all the points of a line segment. Example: the points 1/3 of the distance from either of the the original 2 points are approached but never hit.Please, feel free to rephrase this explanation. I know it's sloppy.
Each point can be connected to every other point. That gives you 10 x 9. However, since this counts each line segment twice, this result should be divided by 2.
You draw a series of line segments joining the points which would be the middle of the top of each bar of the histogram.
The minimum is two points, one point at each end of the line.