Parallel Lines
How will you find the mid points of two parallel lines. You have a ruler without the scale. (Mid points of each of the lines)
Labels: bcreative, mathemagic
posted by Rajesh Lal at
7:41 PM
Monday, November 23, 2009
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31 Comments:
Draw a line at an angle cross both lines with the stright edge, make Plenty long one one side, and mark both places on the straight edge where the lines intersect the straight edge. Slide the straight down on the line you drew until one of the marks are on the other parallel line. Mark a third mark on the straight edge. You now have three equally spaced marks,
Put mark one on the first parallel line, put the third mark on the second parallel line. The middle mark gives you the mid point.
This is another ambiguously worded question.
Ragknot interprets this as the mid-point BETWEEN two parallel lines. I saw it as finding the mid-point ON EACH of two lines which are parallel to each other.
So, which is right?
On my interpretation the two parallel lines would necessarily be of finite length. Let one have points A and B at each end, and the other have points C and D. Assuming the lines are horizontal, A and C are on the left, B and D on the right.
Let X be the intersection of AD and BC. Let Y be the intersection of AC and BD. Continue the line XY through both parallel lines and they cross at their mid-points.
This assumes that AB and CD are of unequal lengths.
If they are equal, then mark your straight edge with the length of AC, then move the straight edge so that it over X (X as defined above). Where your marks sit over each parallel line are their mid-points.
A problem:
Either point x or point y cannot exist as the lines would not intersect:
.A________B
..\ |
...\ |
....\ |
.....\ |
......\ |
.......C__D
Sorry if I did not make myself clear.
X, the intersection of AD and BC, lies between the two parallel lines.
Y is where the extensions of lines AC and BD intersect. Referring to the diagram in the previous post, if CD is the shorter of the two parallel lines then Y will be below CD. IF AB is the shorter one then Y will be above AB.
Join X and Y and extend this line so that it cuts through the two parallel lines. It will do so at the mid-points of AB and CD.
Assumption:
the straight edge of the ruler is like most rulers, in that a line can be drawn at a right angle from another line by placing the small edge of the ruler on the line and the long edge will then be perpendicular to the small edge.
This procedure will find the midpoint of any line (parallel line not required). This procedure can be repeated for each line.
Start at point A on AB (AB is a line from point A to B) and mark off a point at an arbitrary, but conveniently small, distance from point A on line AB. Call this point C. AC is our unit length. Aligning the edge of the ruler at point A and along the line AB then mark the ruler at point C. Unit length is now marked on the ruler.
Now align the edge of the ruler at point C and mark the line at the unit length marking on the ruler. This is a point on the line twice the distance from A as point C is to A. Call this point D. AD is the double unit length. Realign the edge of the ruler to point A, and mark of the ruler at point D. Double unit length is now marked on the ruler.
Now draw a line perpendicular to the line AB at point C (see assumption for explanation). Call this line CE. Now placing the corner edge of the ruler at point A, rotate the ruler until the double unit length marking intersects the perpendicular line CE. Call this point F. Draw the line AF.
The angle between AF and AB is 60 degrees. Since sin= o/h=1/2 , and sin inverse 1/2=60 degrees.
Now mark of the a unit length point on the opposite end of the line. Call this point G. BG is a unit length line. Now draw a line perpendicular to the line AB at point G (see assumption for explanation). Call this line GH. Now placing the corner edge of the ruler at point B, rotate the ruler until the double unit length marking intersects the perpendicular line GH. Call this point J. Draw the line BJ.
The angle between BJ and AB is 60 degrees. Since sin= o/h=1/2 , and sin inverse 1/2=60 degrees.
Extend BJ and AF and, call their intersection point K. Now draw a line perpendicular to the line AB intersecting point K (slide the small edge of the ruler along AB until the long edge intersects K). The intersection of this line and AB is the midpoint of AB.
Why does this work ? It, is because we have constructed an equilateral triangle, where each corner is opposite the midpoint of each side.
Cam
Correction to my above post:
" Since sin= o/h=1/2 , and sin inverse 1/2=60 degrees. "
Should be:
" Since cos= a/h=1/2 , and cos inverse 1/2=60 degrees. "
Cam
test
Assuming the parallel lines are of finite, yet different lengths, here's a solution(if the lines are of identical length, this solution is invalid):
2 parallel lines: AB and CD
Draw lines AD and BC ... they intersect at point F
Starting at A, draw a line extending through C until it intersects a line drawn from B extending through D ... point of intersection is F.
Starting a F, draw a line extending through E until it intersects AB.
The last drawn line intersects AB and CD at their respective mid-points.
CORRECTION to previous post:
Lines AD and BC intersect at point E.
Sorry about that ....
In my solution above, lines AD and BC intersect at point E (not F).
Sorry ......
For either line, or any line you can find its mid point.
Line up one end of the ruler with the line end point. Mark a spot on the ruler that is a more that half the length of the line.
Rotate the ruler to construct intersecting arcs from each end of the line. You may have to make a couple of trials to find the correct intersections. This is approximatly like taking a compass and drawing circles around each end of the line with a radius just more than half the line length.
From the intersections of the arcs, draw a line that will cross the first line at the line's mid point, and the line will also be perpendicular to the original line.
To Zaux - what's the difference between your solution and the earlier one by Wizard of Oz?
See a picture of my solution at
ragknot.blogspot.com
This question is impossible to answer without making many assumptions.
First, the question says "two parallel lines"; by definition a line is infinate in length, therefore, every point on the line is equidistant each "end" therefore in this case, any point you pick is the center (midpoint) of the line.
Second: IF the question presumes the lines are actually line segments, then we need to know if they begin and/or end in the same plane (i.e. Is the endpoint of line segment A 90 deg. from the endpoint of line segment B).
Is your straight edge longer or shorter than the line segments or the seperation distance?
Since this question requires too many assumptions, I take the litteral viewpoint and say that the lines are infinate in length and therefore any point on the line can be the midpoint of that line.
Hey Ragknot,
Some food for thought......
I thought of using the method that you mentioned, as well, but I came to the conclusion that using the ruler to draw an arc like a compass would not be accurate enough, or at least, the accuracy of the arcs would be questionable. It could, of course, be said that the method I mentioned has the potential to add error, as well, (and relies on an accurate right angle between the small and long edge of the ruler), but I believe the error involved would be fairly small. Overall, I believe that, Wiz's method is the most accurate, however it requires two parallel lines.
I suppose, that in an effort to control the inherent errors in the method that you mentioned, one could draw multiple intersecting arcs. Likewise, in the method I mentioned, one could choose multiple unit lengths, in an attempt, to improve the accuracy of the equilateral triangle construction.
Cam
Cam,
You are right. I suppose almost any method will have certain inaccuracies. Laying a ruler across the lines and drawing a line can't be done with 100% accuracy. If you don't have a scaled ruler, then you are hanicapped to begin with.
I was a draftsman for years and with even with years of experience and practice.... well perfection and being human do not often go together. It all depends on what's acceptiable accuracy.
Cam stated above that "Overall, I believe that, Wiz's method is the most accurate, however it requires two parallel lines".
No problem, Cam. If you only have one line then just line up one edge of your ruler with that line and draw a parallel line with the other edge.
In fact, this problem could have been apparently more difficult in the form: Find the mid-point of a finite line using only an unmarked straight ruler.
there is no midpoint of a line. it goes on forever
That's why I said "finite line", i.e. a line that begins at point A and ends at point B.
According to Euclid a line (which may be curved) is a breadthless length and a straight line is a line which lies evenly on itself. i.e. a line most definitely can have a finite length.
A straightedge is not like an umarked ruler and it cannot be treated as a rectangle and cannot be used as a T-square or a compass.
You cannot slide the straight edge parallel to itself (except along the line that it defines) as there is no geometrical construction that lets you do that (without a compass).
Geometry is a branch mathematics where you are supposed to understand that the thickness of the pencil and other such mundane matters are not of concern. Geometry is mathematically exact. Actually drawing a geometrical construction is only done to illustrate the underlying maths (when coupled with e.g. Euclids axioms).
You are not allowed to make an infinite sequence of successive approximations.
Wiz's treatment of two equal length lines is not geometrically valid. The best I can think of is to increase (or decrease) the length of one of the lines equally at both ends so that it has the same midpoint but a different length. That could be done by copying an arbitrary increase in length at one end to the other by marking on the straightedge - but I doubt that that is a valid geometrical operation with a straight edge.
That's my 2 cents.
Fair comment, Chris.
With two parallel lines of equal length (AB and CD) there is a "geometric" solution to finding their mid-points: draw another line of different length parallel to these (using your straight ruler, assuming that it has parallel edges) and use this as described earlier against AB and CD in turn to find their mid-points.
Alternatively, pick any point inside CD, call it M, and use the line CM to find the mid-point of AB. Likewise pick another point inside AB to find the mid-point of CD.
a true line does not have a midpoint,because it has not end point but rather goes on forever in either direction.
Last Anonymous; lines don't have to be infinite.
Wiz, a straightedge only one has one edge. You cannot draw a line parallel to a given line using only a straightedge. Your last argument about making an arbitrary point on each line and using it to find the midpoint of the other line is brilliant; I bow to you.
Draw an "X" at opposite lines...you'll get the midpoint...
Prof Em ... your solution determines a mid-point between the lines, rather than a mid-point of each line. The problem states the search is for mid-point"S" (plural) of the lines. Your suggestion is the first step in geometrically determining the mid-points of the 2 lines, provided the lines are finite and of different lengths.
I really don't understand why some of you think that a line has to be infinite - that belief is just wrong. It's why we say "infinite line" rather than plain "line". The definition of a line was made by Euclid.
Every point on an infinite line is a mid point.
well you technically could not find the mid point of two infinite lines due to the fact that each time you chose different end points to plug into the mid point formula your mid point would constantly change
Thee is no "the midpoint" of an infinite line.
I'd calculate the midpoint is (∞ + (-∞))/2 = ∞
Hey! It never said you can use a pencil, just an unmarked ruler!
Hi steelersfan. It is frequently the case that the pencil is taken for granted. The classical geometers only used a straightedge and compass - they always neglect to mention the pencil.
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