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Math Forum / Mathematics / General Topics / July 2009Parallel Lines
How do I prove 2 straight lines are parallel to each other? If they intersect then they are not parallel. But how do I prove they are parallel? Thanks > How do I prove 2 straight lines are parallel to each other? If they > intersect then they are not parallel. But how do I prove they are > parallel? > > Thanks If the alternate angles they make with a transversal are equal...  SignatureVirgil Virgil wrote on 28-06-2009 : >> How do I prove 2 straight lines are parallel to each other? If they >> intersect then they are not parallel. But how do I prove they are [quoted text clipped - 3 lines] > > If the alternate angles they make with a transversal are equal... How do I know the alternate angles are equal? I do not want to use any geometry box instruments? Thanks > Virgil wrote on 28-06-2009 : > > [quoted text clipped - 10 lines] > > Thanks If you are unable to make any comparisons at all, then your "geometry" cannot distinguish between parallel and skew for line segments which do not actually intersect. SignatureVirgil >How do I prove 2 straight lines are parallel to each other? If they >intersect then they are not parallel. But how do I prove they are >parallel? There must be some information regarding the lines that would let you prove (or disprove) the assertion. If not, this becomes a problem with insufficient information. SignatureRemove del for email Barry Schwarz wrote on 29-06-2009 : >> How do I prove 2 straight lines are parallel to each other? If they >> intersect then they are not parallel. But how do I prove they are [quoted text clipped - 3 lines] > prove (or disprove) the assertion. If not, this becomes a problem > with insufficient information. I am just given 2 segments of lines. Now, if I can draw perpendiculars on the both the line from certain points from the other lines and if I measure the length of the perpendiculars with rulers or something and they are found to be same then they are parallel. Am I right? But without the ruler, how do I prove they are parallel to each other, geometrically? Thanks > Barry Schwarz wrote on 29-06-2009 : >> [quoted text clipped - 13 lines] > > Thanks You can't just draw lines arbitrarily and use measurements to prove two lines are parallel. "Parallel" is an abstraction, and must be proved in abstract way. If you could prove that the perpendiculars of two straight line segments were themselves perpendicular, then you could prove the original lines were perpendicular. However, it doesn't seem to help much, as you have the same problem. If you know the co-ordinates of points on the line segments, then it is simple algebra to get the formula for the lines and see if they are parallel, in two, three or however many dimensions you may wish. Peter Webb wrote on 29-06-2009 : >> Barry Schwarz wrote on 29-06-2009 : >>> [quoted text clipped - 26 lines] > simple algebra to get the formula for the lines and see if they are parallel, > in two, three or however many dimensions you may wish. Let's say, I have 2 straight line segments AB and CD. Now if I draw a perpendicular from point C on AB (let's say at the point E on AB) and another perpendicular from point B on CD (let's say at the point F on CD), then the angles CEB and ABF are both right angles and opposite to each other. So ECFB is a rectangle. So AB and CD are parallel to each other. Is it right? > Peter Webb wrote on 29-06-2009 : > [quoted text clipped - 37 lines] > > Is it right? Do you know theses points A,B,C and D? If not, so the problem is unsolvable because you don´t have any information about the lines. But, if you know the coordinates of the points so what you have to do is to find the equation of the two lines and compare the slopes. Different slopes implies not parallel. This solve the problem for the particular case of R^2, but for if your plane is the projective plane, for example, the answer is, NEVER. (Sorry my bad english) > Barry Schwarz wrote on 29-06-2009 : >> [quoted text clipped - 7 lines] > > I am just given 2 segments of lines. in what form are these linesegments presented to you? > Now, if I can draw perpendiculars > on the both the line from certain points from the other lines and if I > measure the length of the perpendiculars with rulers or something and > they are found to be same then they are parallel. Am I right? But > without the ruler, how do I prove they are parallel to each other, > geometrically? Jasen Betts formulated the question : >> Barry Schwarz wrote on 29-06-2009 : >>> [quoted text clipped - 9 lines] > > in what form are these linesegments presented to you? In a two dimensional plane white paper >> Now, if I can draw perpendiculars >> on the both the line from certain points from the other lines and if I >> measure the length of the perpendiculars with rulers or something and >> they are found to be same then they are parallel. Am I right? But >> without the ruler, how do I prove they are parallel to each other, >> geometrically? > Jasen Betts formulated the question : > >> Barry Schwarz wrote on 29-06-2009 : [quoted text clipped - 12 lines] > > In a two dimensional plane white paper In that case, the problem is unsolvable. SignatureWhich of the seven heavens / Was responsible her smile / Wouldn't be sure but attested / That, whoever it was, a god / Worth kneeling-to for a while / Had tabernacled and rested. > Jasen Betts formulated the question : > >> Barry Schwarz wrote on 29-06-2009 : [quoted text clipped - 6 lines] > > In a two dimensional plane white paper .... Perhaps you just want a simple practical test. If so, your word "prove" has been misleading people. Theoretical geometry has straight lines of zero thickness, perfect parallels, etc. which can have things proved about them. However, you may actually want to check whether your two physical lines drawn on paper are nearly enough parallel for practical purposes. In that case, I suggest using a ruler and set-square. Lay one side of the set-square along your first line, put the ruler firmly against another side, then slide the set-square along the ruler to your second line and see whether it fits. Ken Pledger. Ken Pledger presented the following explanation : >> Jasen Betts formulated the question : >>>> Barry Schwarz wrote on 29-06-2009 : [quoted text clipped - 20 lines] > > Ken Pledger. "...see whether it fits" - there is my problem. I do not wish to see or measure because that could lead to errors. My eyes, rulers, set-square or whatever instruments could be defective and could produce erroneous result. I want to prove it geometrically. I know how to draw a perpendicular from a given point. So as I earlier said, - if I am given 2 segment of lines on a plain paper, and if I can draw perpendiculars from each line onto the other line - then the opposite angles will be right angle. So that will become a rectangular. So the given line segments will be parallel to each other. Is that right? Thanks JC wrote : > Ken Pledger presented the following explanation : >> [quoted text clipped - 37 lines] > > Thanks You don't seem to understand what all others already said : It is impossible Because you have no information on the given lines : they are just two physical segments on physical paper sheet, drawn in an unknown way with a physical pencil. This doesn't allow to proove anything at all. From no information you positively can't extract some information ! Hence you NEED to measure something on your physical sheet of physical paper, and deduce "Parallel at x% accuracy". Or to know HOW these lines have been drawn = constructed. Regarding your "method" it however has a big flaw : <http://cjoint.com/?gEjCEt3dX2> Do you really think that because angles B and D are right angles then ABDC is a rectangle ??? (and then that AB is parallel to CD) Regards. SignaturePhilippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) > JC wrote : >> Ken Pledger presented the following explanation : [quoted text clipped - 62 lines] > > Regards. Thanks for the explanation. Now I understand. I have to think harder. > JC wrote : >> Ken Pledger presented the following explanation : [quoted text clipped - 62 lines] > > Regards. Well, that's accepted. Now once again, let's say there are 2 segments of straight lines AB and CD. I draw a perpendicular from point B onto CD which intersects at, say, E. Now, from E on CD I draw another perpendicular EF. If EF and CD are same line then AB and CD are parallel. Is it right? Thanks > > JC wrote : > >> Ken Pledger presented the following explanation : [quoted text clipped - 68 lines] > intersects at, say, E. Now, from E on CD I draw another perpendicular EF. > If EF and CD are same line How do you know they're the same line? What if they're a billionth of an inch apart or have a 1 part per billion different slope? > then AB and CD are parallel. > > Is it right? > Thanks Mensanator laid this down on his screen : >>> JC wrote : >>>> Ken Pledger presented the following explanation : [quoted text clipped - 63 lines] > a billionth of an inch apart or have a 1 part per billion > different slope? What I am trying to say is theoretically they are parallel. If they are even a billionth of an inch apart, then they are unparallel. If I can measure a billionth of an inch with some instrument, then I could use the same instrument to see whether they are not a billionth of an inch apart. >> then AB and CD are parallel. >> >> Is it right? >> Thanks [ summarized all that horrible Google like quoting ] JC wrote : > Mensanator wrote : >> JC wrote: [quoted text clipped - 17 lines] > the same instrument to see whether they are not a billionth of an inch > apart. So the only thing you can say is "they are parallel at a 1b. of an inch accuracy" You can't then know if they are really parallel or not. Theoretically parrallel is an abstraction that you definitely _can't measure_ "Theoretically" means that you can proove it by pure logical deduction from other previously known properties of your lines. Not by measuring anything, even with an unusual very accurate instrument able to measure less than diameter of quarks, or angle of less than a quark seen from the border of universe. Regards. SignaturePhilippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) Philippe 92 laid this down on his screen : > [ summarized all that horrible Google like quoting ] > [quoted text clipped - 36 lines] > > Regards. But that way we cannot say anything perfect. When we draw a circle, if we could measure with that instrument, we would find there is a small difference where the point finally meets at the starting point. > Philippe 92 laid this down on his screen : > [quoted text clipped - 39 lines] > > But that way we cannot say anything perfect. Only when you draw it, or try to measure drawn objects. > When we draw a circle, if > we could measure with that instrument, we would find there is a small > difference where the point finally meets at the starting point. Let me give you an example of where things ARE perfect. I have an equation: g = (Xa - Z)/Y Where 'a' is a variable and X,Y & Z are constants. Although I don't know what the values of X,Y & Z are, I DO know that X is a power of 2 and Y is a power of 3. That equation is that of a straight line whose slope is X/Y and intercept is -Z/Y. From this, it follows that the slope cannot be 1 and therefore MUST intersect the identity line f(a) = a whose slope is exactly 1. And this intersection will be the rational number Z/(X-Y). And this is true even though I don't actually know what X,Y & Z are. If I ever do find that out, I simply plug in those values and I know exactly the point where the two lines intersect. Absolute perfection. No pencils were harmed in this production. Mensanator wrote on 04-07-2009 : >> Philippe 92 laid this down on his screen : >> [quoted text clipped - 67 lines] > > No pencils were harmed in this production. Ok, let me frame my explanation in a little different way. Let's consider 3 different points on the straight line CD - as I earlier mentioned - namely, X, Y and Z. Now, if the new perpendicular EF from the point E on CD intersects at those 3 points X,Y,Z, - then EF and CD have to be the same line. Since in a 2-dimensional plane, if a straight line intersects another sraight line at more than one different point, then both the straight lines are same. > Mensanator wrote on 04-07-2009 : > [quoted text clipped - 73 lines] > consider 3 different points on the straight line CD - as I earlier > mentioned - namely, X, Y and Z. Now, > if the new perpendicular EF from > the point E on CD intersects at those 3 points X,Y,Z, False premise. A perpendicular of a line can only intersect at a single point. > - then EF and CD have to be the same line. A perpendicular to CD is CD? Better re-think that. > Since in a 2-dimensional plane, if a straight > line intersects another sraight line at more than one different point, > then both the straight lines are same. But not perpendicular. Mensanator formulated on Saturday : >> Mensanator wrote on 04-07-2009 : >> [quoted text clipped - 81 lines] > > But not perpendicular. You forgot my earlier post. :-) My earlier post says - "... let's say there are 2 segments of straight lines AB and CD. I draw a perpendicular from point B onto CD which intersects at, say, E. Now, from E on CD I draw another perpendicular EF. If EF and CD are same line then AB and CD are parallel." I didn't say, at that time, what I meant by "same". In my last post I tried to do that. > Mensanator formulated on Saturday : > [quoted text clipped - 93 lines] > I didn't say, at that time, what I meant by "same". In my last post I > tried to do that. Ok, forget all that then. Now, if the lines in question are mathematically accurate, they have 0 thickness and are thus, invisible. How can you construct a perpendicular to something that's invisible? The fact that you can see the lines implies they are not mathematically accurate, thus, youe proof fails before you even pick up your compass. Mensanator wrote : >> Mensanator formulated on Saturday : >> [quoted text clipped - 97 lines] > are not mathematically accurate, thus, youe proof > fails before you even pick up your compass. Why should they have 0 thickness? all the lines are drawn with same pencil and instruments. They should be having equal thickness, whatever that may be. > Mensanator wrote : > [quoted text clipped - 101 lines] > > Why should they have 0 thickness? Because they're defined that way: http://en.wikipedia.org/wiki/Line_(geometry) Lines are an idealisation of such objects and have no width or height at all and are usually considered to be infinitely long. http://mathworld.wolfram.com/Line.html A line is a straight one-dimensional figure having no thickness and extending infinitely in both directions. > all the lines are drawn with same > pencil and instruments. They should be having equal thickness, whatever > that may be. And, therefore, are not lines in the mathematical sense. Such drawings are abstractions used to represent lines. They are useful up to a point. What you're trying to do is beyond that point. You can do absolute proofs using the algebra of lines (if you know it) as I showed in my example. But with geometry, you are restricted to the limitation of your representations Mensanator wrote : > You can do absolute proofs using the algebra of > lines (if you know it) as I showed in my example. No. Because the coordinates of your points do not come from scratch Either you PREVIOUSLY know them (that is the lines have be draw FROM two pairs of previously known points, Or you MEASURE them. > But with geometry, you are restricted to the > limitation of your representations No. Restricted to your deductive abilities. Exactly equivallent. If you don't know anything in advance about your lines, you can't deduce anything. Your are just able to measure things. No more, no less than the algebraic method. Regards. SignaturePhilippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) > How do I prove 2 straight lines are parallel to each other? If they > intersect then they are not parallel. But how do I prove they are > parallel? > > Thanks same slope, different value at x=0 (or anywhere else) |
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