
#1
Aug110, 02:36 AM

P: 6

So for a while i have been wondering if there was a way to find a point a certain distance along a linear function, so I decided that with my extreme precalc level of math that I would try and write an equation for it. long story short I would appreciate it if someone would take a look at the equation and try and figure out where I goofed up. The equation I came up with is:
c = sqrt((ox+a)^2+(m*(ox+a)+boy)^2) where c is the distance of the hypotenuse of a triangle with coords (ox,oy),(ox+a),(ox+a,f(ox+a)) a is the distance of the horizontal leg of the triangle m, b are the slope and yintercept of the line respectively ox, oy are the original x and y coords on the line px, py are the projected coords along the line z is the distance along the line to the new projected point deltax = (a*z)/c where deltax is the distance to the new point along the xaxis px = deltax + ox py = m(deltax+ox)+b where px, py are the projected coords along the line I hope this post isn't too incomprehensible seeing as I'm writing this at 12:30 



#2
Aug110, 02:53 AM

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I have to admit you should have written it out at another time
If you let your starting point be [tex](x_o,y_o)[/tex] on your line [tex]y=mx+b[/tex] and a distance d along this line to the new point [tex](x,y)[/tex] then you have equations: [tex]d=\sqrt{(xx_o)^2+(yy_o)^2}[/tex] by the distance formula [tex]m=\frac{yy_o}{xx_o}[/tex] for the gradient formula Now all you need to do is solve these two equations simultaneously for x and y (m, d, y_{o}, x_{o} are all constants you know). 



#3
Aug110, 03:39 AM

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Note that from the second equation Mentallic gave you,
[tex](yy_{0})=m(xx_{0})[/tex] Use this in your first equation to solve for x (you will get 2 solutions) 



#4
Aug110, 06:56 AM

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Formula for finding point on a line given distance along a line
Essentially what you are doing is finding the intersections of the line [itex]y y_0= m(x x_0)[/itex] and the circle [itex](xx_0)^2+ (y y_0)^2= c^2[/itex].
Replacing y in the circle equation by [itex]y_0+ m(x x_0)[/itex] gives the quadratic equation [itex](x x_0)^2+ m^2(x x_0)^2= c^2[/itex] which can be factored as [itex](x x_0)^2(1+ m^2)= c^2[/itex] and so [itex](x x_0)^2= c^2/(1+m^2)[/itex] and, finally, [tex]x= x_0+ \frac{c}{\sqrt{1+m^2}}[/tex] 



#5
Aug110, 07:18 AM

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[tex]x=x_0+\frac{c}{\sqrt{1+m^2}} , x>x_0[/tex] [tex]x=x_0\frac{c}{\sqrt{1+m^2}} , x<x_0[/tex] since c is the distance (a positive value) and [tex]\sqrt{1+m^2}[/tex] is also positive for all real m. 



#6
Aug110, 09:22 AM

P: 181

I don't know if I read the meaning of your constants and variables right, but it seems like the equation should be:
c = sqrt((a)^2+(m*(ox+a)+boy)^2) instead of: c = sqrt((ox+a)^2+(m*(ox+a)+boy)^2) If all you're doing is using the Pythagorean theorum, and a is just the horizontal leg of the triangle. PS try Latex in your posts all you have to do is go into advanced reply mode, and use the tags [tex] and [/tex] at the start and end of your equation. It makes things look pretty! 



#7
Aug110, 09:44 AM

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Whether what he did was correct or not so far, the fact is that he wants to find the point that is a certain distance from some other point along a straight line. This implies that he wants coordinates (x,y) in terms of d=distance, m=gradient, (x_{o},y_{o})=initial point. And I'm just not seeing that.
p.s. I gave the the OP's work another little glance and I'm looking away again. I'm simply afraid of delving into it... And it would have been more appealing to read if you had used subscripts. the intial point (x_{o},y_{o}) shouldn't be expressed as (ox,oy) since this suggests o times x and o times y. Then we wonder what the hell o is. 



#8
Aug110, 09:51 AM

P: 181

Chill, we were all in high/middle school once I don't know of that many people who natively speak Latex. I only just learned how to use it myself a few weeks ago.
I think my correction may be on the right track OP, would you mind clarifying what c and z are exactly? I think of c as just the hypotenuse/distance. 



#9
Aug110, 01:11 PM

P: 6

So heres the equation with wil3's correction (with latex =D)
c = [tex]\sqrt{((a)^2+(m*(x_o+a)+by_o)^2)}[/tex] a, and c exist to establish legs of a triangle to get a ratio of line length to horizontal length, and yes I found this using the Pythagorean theorem. deltax is the horizontal distance from [tex]x_{o}[/tex] to [tex]p_{x}[/tex], whereas a is the arbitrarily set value to use in the Pythagorean theorem. And yes I suppose you could do this using a circle and whatnot. 



#10
Aug110, 08:43 PM

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#11
Aug110, 10:39 PM

P: 6

No, c isn't the distance along the line, c is the hypotenuse of the triangle that is used to find the ratio of the x length to the distance along the line. Then using this ratio:
deltax the total change in x c the hypotenuse of the triangle a the horizontal leg of the triangle z the distance to travel along the line deltax/z = a/c deltax = az/c 



#12
Aug110, 11:44 PM

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P: 3,436

Oh I see what you mean now, but I still don't understand why you would need to do that. Why make a small right triangle with hypotenuse c and horizontal length x when the distance between your starting and ending point is a larger right triangle with hypotenuse z and horizontal length [itex]\Delta x[/itex]. Sure you can find a relationship between the small and larger triangles by similarity, but what's the point? You've just introduced more variables which aren't necessary and just make things more cumbersome.
Is the answer [tex] \sqrt{((a)^2+(m*(x_o+a)+by_o)^2)} [/tex] what you were looking for? Hallsofivy and I have pretty much given you the answer to your question (at least I think we did), it just depends on what your question actually was. 



#13
Aug210, 06:21 AM

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#14
Aug210, 08:57 AM

P: 6

C does lie on the line, however it is not the total distance to the point.




#15
Aug210, 08:59 AM

P: 181

Would you mind posting a picture or diagram with all the variables/constants labeled? This thread has proven to be a lot more interesting than I expected, but I know that at least I am a little bit confused as to the specific meaning of some of your variables. Thank you very much!




#16
Aug210, 09:50 AM

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P: 3,436

[tex]\frac{\Delta x}{z} = \frac{a}{c}[/tex] Doesn't tell us anything useful, since we're comparing our required sides of the triangle on the left side with a redundant triangle. Would you mind please taking a look at my previous post and specifically this part, 



#17
Aug210, 09:55 AM

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This is how I interpreted the variables. Note that the question the OP asked implies that we are searching for the point (px,py) given we know (ox,oy), y=mx+b and z.
Uploaded with ImageShack.us 



#18
Aug310, 12:21 AM

P: 6

Yeah, thats it I guess you've been right that I didn't have to do that first ratio with the a and c. I never really saw it like that, thanks.



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