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Math Help - Using only a point and slope

  1. #1
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    Question Using only a point and slope

    Given that I know the value of a single point (x1,y1) on a line and its splope m, what formula(s) can I use to determine the two other points that are exactly d units from the known points.
    This is for use in a server side web application, written in php.
    At 60 yrs old, I have forgotten all that I thought I knew about doing this
    That includes trying to solve for two variables in a set of equations, which I suspect this will need

    TIA,
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  2. #2
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    point slope formula:
    (y-y_1) = m(x-x_1)

    you can derive this from definition of a slope by dividing both sides by (x-x_1)


    if you know distance d, then use the distance formula: d = \sqrt{(x_1 -x_2)^2 + (y_1 - y_2)^2}
    substitute your known points for (x_1, y_1)
    then solve the equation for (x_2,y_2)

    It is easier if you know trig. d_x = d * \cos \theta \mbox{ where } d_x \mbox{ is the x component of the distance}

     \theta = \tan^{-1}(\mbox{slope})

    solve for d_x

    x_2 = x_1 + d_x
    or
    x_2 = x_1 - d_x
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  3. #3
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    Red face

    I will have to 'absorb' what you posted a bit!

    Like I said, solving equations with two variables (x2 and y2) is not my strong point, and more importantly, I will have to do this programatically, not on paper.

    Let me ponder
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  4. #4
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    Using the trig approach seems a bit easier but .....

    Dx = (cos(arctan(slope))/d

    so x2 = x1 + Dx

    then do I just stick in x2 into (y2-y1)=slope(x2-x1) and solve for y2 ?

    Is it just that easy!!!
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  5. #5
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    Relative to my previous message, what is the trig for solving for dy first?
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  6. #6
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    Unhappy Problem reworded.

    Still having some difficulty, so perhaps a better explanation of what I am doing is in order.

    I am plotting co-ordinates for an airfoil cross section (to make ribs) from a database of many airfoils. These points are 'scaled' to be exactly 1 unit from Leading Edge to Trailing Edge, and I am plotting them by drawing into a blank .png file (jpg or tiff would also work). In order to get them sized to a usable size, I am scaling them up by multiplying by the desired number of inches for the chord (LE to TE) and by the number of pixels per inch on output. As a result, the graph is made up of many short straight lines. This is all working fine for me so far.

    But many of these wings are covered (or partly covered) by a 1/16 inch or 3/32 inch sheeting, which requires that the rib is plotted that much smaller and inside the original.

    My intent was to match each line segment with a parallel one that is exactly 1/16 or 3/32 inch 'inside' the original. So, that is why I started with the slope of the original line, slope of a perpendicular line, and one of the original end (or start) points. From there I should be able to calculated the inside point along the perpendicular slope at the desired distance (1/16 or 3/32 inch).

    If all of this still makes sense to anyone , could you help me. My first attempts at doing this were good for the line segments that had a reasonable vertical dy value, but would colapse elswhere.
    Attached Thumbnails Attached Thumbnails Using only a point and slope-clarky.png  
    Last edited by griffinmt; November 18th 2009 at 09:28 AM. Reason: Add .png file
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