Are the set of all centre points ç of XçY
a) in a straight line (i.e. colinear), and
b) a reasonable approximation of the centre of the path bounded by AB and CD
Well, if the two line are perpindicular then the closest point on CD to any random point on AB would be the point of intersection, therefore the same point for all points on AB. This is easy to prove to yourself if you just think about, take any point on AB and the point of intersection, and draw a circle with the AB point as the center. All points on CD must lie outside the circle except the point of intersection and are therefore further away.
I’m too lazy to try to prove it, but it seems to me that the answer is obviously yes. Wouldn’t the set of the centre points of XY form the line that bisects the smallest angle between AB and CD (if AB and CD are neither perpendicular nor parallel?) If AB and CD are either perpendicular or parallel the cases become trivial.
If the two lines aren’t parallel, one can translate and rotate the reference frame such that the CD is the x-axis and AB intersects CD at the origin. Algebraically, AB has formula y=nx. (except for the other trivial case, where AB and CD are perpendicular.) The shortest line segment from X to CD must be perpendicular to CD, so for a given point on AB (y,ny) this midpoint will be at (y,(ny/2)); the formula for the (set of all center points) line is thus y=nx/2.
Long story short: Arnold is correct; it’s the bisector of the smaller angle.
1a is true for all cases. 1b is true for all cases except the perpendicular one, in which the new line is identical to AB.
If I understand what you are asking, yes. I’m assuming you really mean that AB and CD are infinite lines, and not line segments between two endpoints. In that case, the shortest segment from any given X point to the other line will be the one that forms a perpendicular with the line (a segment drawn to any other point on the line would form the hypotenuse of a right triangle with the perpendicular as one of the sides, and have to be longer). As long as the lines themselves aren’t perpendicular, you can always do this. So your set of X-Y segments are a set of perpendiculars to line CD and all parallel.
Now just imagine that you viewed the CD line as the X-axis of a graph. What you are doing when you find the midpoints is simply taking half of the Y-values for the other line, in other words, you are constructing a line with 1/2 the slope, and intersecting the CD line at the same point as the AB line. If what you mean by “path bounded by AB and CD” is the area between the two lines, yes, you are describing the center of that area, and bisecting the smaller angle, as indicated by Arnold.
The two lines being perpendicular is a special case which has already been addressed. If, in your original question, you really meant two line SEGMENTS between points A and B and between points C and D, it gets more problematical because the shortest point for your X point may be a perpendicular, OR it may be a line segment to the nearer endpoint of CD.
And (Tim) and I just simulposted and said essentially the same thing, I see.
Oops. I was mistaken earlier. e.g. the y=50x does not bisect the lines y=100x and y=0. In one sense, it’s halfway between, but the anglular difference between y=50x and y=100x is a lot smaller than that between y=0 and y=50x.
So, 1b is true for parallel lines and is a good approximation when they are near parallel, but if the lines are nearly perpendicular it is a lousy approximation, as the midpoints are noticably closer to AB.
Yes, including 3d spacial coordinates, if lines AB and CD are straight and not curved. A line can be curved, if any of the two lines are curved this does not always hold true. Two non straight lines which are always increasing or decreasing the same amount in opposite directions would also form a straight line at the median of the line segments in 3d spacial coordinates. This is also possible if you go with two lines spiraling. Such as you see demenstrated by a DNA double helix. A staight line would be formed at the median of the line segments.
And we made the same mistake simultaneously, too. My apologies.
Just a nitpick, and it goes back far too many years for me, but if I recall correctly, the notation AB is used for line segments and not lines.
yabob
For this discussion I will only talk about straight lines in two demensions. You seem to be thinking in two demensionial X,Y coordinates, so lets go with that. The only time that the shortest line segment of XY is perpendicular to point X on line AB intersecting line CD is when the two infinate lines run parallel to each other. Suppose that line AB and CD run at 40 degrees to each other. In that case the line segment would have to come off of line AB at point X at 60 degrees for the shortest line segment. Remember that a triangle consists of 180 degrees total for all the interior angles. Both lines have now intersected with line segment XY at 60 degrees. Using the degrees of deviation each line has in two demensions the triangular method I state will give you the shortest line segment XY. The median of the line segments will always give you a straight line that runs half way between lines AB and CD.
ACtually Phobia, you are a little off, if the two lines form an angle of 40 degrees then the corresponding angle off of AB must be 50 degrees. This is because the line segment must in fact be perpindicular off of CD not AB. This is demonstrated once again by drawing circle that is tangent to CD and has the point on AB as its center.
Not perpendicular to line AB - perpendicular to line CD. The shortest segment from a point not on a line to the line will be a perpendicular to that line. and the OP said to pick a point X on line AB, then construct the shortest line segment to CD (perpendicular to CD). Then take the midpoints of all such segments.
If the angle between the lines is 40 degrees, it forms a 40 - 50 right triangle. As (Tim) and I both (eventually) said, this gives you a line with half the slope of AB relative to CD, which does NOT bisect the angle. If it did, arctan(2X) would = 2arctan(X).
A line segment XY is not at it’s shortest length when perpendicular to line AB or line CD. The statement was for point X to be put anywhere on line AB and the shortest line segment that intersected line CD would be a point Y. Continue to randomly place point X in any location on line AB and this holds true.
YOu dream of geometry? Matt, do you have any idea how SICK and DISGUSTING that is? I mean, that’s just evil and perverted!
Yes, perpendicular on line CD is the shortest XY line segment that can be obtained. You still get a linear line for all the median if the lines AB and CD meet the conditions I mentioned earlier.
Yes, including 3d spacial coordinates, if lines AB and CD are straight and not curved. A line can be curved, if any of the two lines are curved this does not always hold true. Two non straight lines which are always increasing or decreasing the same amount in opposite directions would also form a straight line at the median of the line segments in 3d spacial coordinates. This is also possible if you go with two lines spiraling. Such as you see demenstrated by a DNA double helix. A staight line would be formed at the median of the line segments.