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How to find two endpoints of a line?

  1. Aug 11, 2009 #1
    how to find two endpoints of a line..??


    I have a particular point on a curve.. This curve isnt a smooth curve.. anyways, i need to draw a tangent at this point.. i have the following information in hand:

    1. Point P at which tangent has to be drawn.
    2. Length l of the tangent.

    Now i want the two end-points of the tangent in such a way that P is the mid-point of these new points.. I dont have equation of the curve. i dont know slope of the tangent..

    Is there any way to do this.??

    Thanks in advance
  2. jcsd
  3. Aug 11, 2009 #2


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    Welcome to PF!

    Hi rakeshthp! Welcome to PF! :smile:
    Well, what do you have? :confused:

    Can't you just line up a straight-edge against the curve so that it touches it at P?
  4. Aug 11, 2009 #3


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    Re: how to find two endpoints of a line..??

    In what sense, then do you "know the tangent" line? And certainly you need to know the tangent in some way before you can find points on it!
  5. Aug 12, 2009 #4
    Re: how to find two endpoints of a line..??

    Let me be more precise... Pls refer to the image diagram attached along this post..

    You will find 5 points.. I need to calculate points a, b, c and d.. SO i thought i would begin by finding points a, d

    lengths (Pa) = (Pd), is not constant.. it may vary every time.. it is called "along boundary length".. So, for a given along boundary length, i need to find these two points a, and d, with respect to P in such a way that,

    and points a and d must lie on any of the line segments..

    First way i thought was to find the tangent(shown in green), of "along boundary length" and get two end points.. from these two end points i can draw a line which will intersect the arc, and i can get values of "a" and "d"... But now i feel this logic is having some fault..

    Now am stuck up.. My ultimate goal is to find values of points a, b, c and d for given set of inputs:

    1. Point P
    2. along boundary length l(Pa), l(Pd)
    3. off boundary length l(ab), l(cd)

    Any way to get this solved..??

    Thanks in advance

    Attached Files:

    Last edited: Aug 12, 2009
  6. Aug 12, 2009 #5


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    Sorry, I've no idea what this question is about. :confused:

    What are the other 6 points along the line in the diagram?

    What are b and c?

    What is the whole of the original question that this comes from?
  7. Aug 13, 2009 #6
    Re: how to find two endpoints of a line..??

    hi Tim..

    Pardon me for my poor vocabulary.... Here is the problem description in detail

    The line(connected line) drawn is assumed to be a boundary (assume it as ocean boundary or Great wall of china).. and the along-boundary length is the distance between point P and point a & point P and point d..

    off-boundary length means the distance away from the boundary.. in this case distance between point p and mid point of line bc..

    Actually, the flow goes something like this.. In this system:

    1) user will first input the points of the arc(connected line).. Using these points, connected line is drawn.. Among these points one point is point P..

    2) Next user will define a region to perform some operation. This region is defined by

    i) Inputting the point of interest around which the region is defined. Here in our example, point P is the point of interest, where user wants to define a region.
    ii) Inputting the along-boundary length. Using this length, we need to find the points A and D..
    iii) Inputting the off-boundary length.. Once Points A and D are estimated, next step is to estimate, points B and C in such a way that distance of point P and mid-point(BC) is equal to the off-boundary length. And these two points need to be perpendicular to point of interest P.

    3) After estimating all four points, users region is defined. And this region has to be in a rectangular shape.

    4) Once the area is defined, further processing is done..

    Any more queries, then pls feel free to ask.. I hope this information makes my problem understandable.....
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