Getting P3 from P1, P2. Dist P3-P1 > Dist P1-P2 & P3 on line P1-P2

Math and Physics Programming

Started by homer_3 April 08, 2011 01:12 AM

3 comments, last by Zakwayda 13 years ago

173

Author

April 08, 2011 01:12 AM

I'm trying to get point 3 from points 1 and 2 such that point 3 is on the same line that points 1 and 2 form, the distance between P1 and P3 is greater than P1 and P2, and the vector direction from P1 to P3 is the same as P1 to P2. This is what I have, but it doesn't work at all. Can anyone point out where I went wrong?

Origin is top left. +X is right. +Y is down.





lv_destPos = OrderGetTargetPoint(EventUnitOrder());

lv_destX = PointGetX(lv_destPos);

lv_destY = PointGetY(lv_destPos);



lv_curX = PointGetX(lv_curPos);

lv_curY = PointGetY(lv_curPos);



//get slope

lv_m = (lv_curY-lv_destY)/(lv_curX-lv_destX);



//get y intercept

lv_b = lv_y - (lv_m * lv_x);



//clicked to right

if(lv_destX > lv_curX)

{

    lv_destX = lv_destX + 50.0;

    lv_destY  = (lv_m * lv_destX) + lv_b;

}else

{

    //clicked to left

    if(lv_destX < lv_curX)

    {

        lv_destX = lv_destX - 50.0;

        lv_destY  = (lv_m * lv_destX) + lv_b;

    }else

    {

        //clicked below

        if(lv_destY > lv_curY)

        {

            lv_destY = lv_destY+50;

            lv_destX = (lv_destY - lv_b)/lv_m;

        }else

        {

            //clicked above

            lv_destY = lv_destY-50;

            lv_destX = (lv_destY - lv_b)/lv_m;

        }

    }

}



lv_destPos = Point(lv_destX, lv_destY);

Emergent

982

April 08, 2011 02:57 AM

Any point on the line containing points p1 and p2 can be written as,

p = (1 - t) p1 + t p2

for some real number t. When t=0, you get p1. When t=1, you get p2. When 0 < t < 1, you get a point in-between (e.g., when t=1/2, you get the midpoint). When t is negative, or when it is greater than 1, you get a point beyond p1 or p2 on the line, respectively.

alvaro

21,604

April 08, 2011 12:33 PM

What Emergent said. Maybe you'll find it easier to think of that formula as
P3 = P1 + t*(P2-P1)

Plugging in t=2 will give you a P3 that satisfies everything you asked for.

homer_3

173

Author

April 08, 2011 11:10 PM

[color=#1C2837][size=2]P3 = P1 + t*(P2-P1)[/quote]
[color=#1C2837][size=2]

[color=#1C2837][size=2]I remember doing this before and I thought it was P1 * t*(P2-P1). But that didn't work for me so I tried just getting the line formula and going from there. I'll try this out though, thanks. Just to be clear
[color=#1C2837][size=2]

[color=#1C2837][size=2]P3.X = P1.X + t*(P2.X-P1.X)
[color=#1C2837][size=2]P3.Y = P1.Y + t*(P2.Y-P1.Y)
[color=#1C2837][size=2]

[color=#1C2837][size=2]correct?

Zakwayda

2,389

April 09, 2011 01:57 AM

[[color="#1C2837"]P3.X = P1.X + t*(P2.X-P1.X)
[color="#1C2837"]P3.Y = P1.Y + t*(P2.Y-P1.Y)

[color="#1C2837"]correct?

As far as the parametric line form goes, yes, that looks correct.

Getting P3 from P1, P2. Dist P3-P1 > Dist P1-P2 & P3 on line P1-P2

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