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Find point between 2 points?


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#1   Members   -  Reputation: 103

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Posted 17 April 2011 - 05:54 PM

I am having a problem finding a point at a specific distance between 2 points. Suppose I have point A(1,1) and point B(4,4). I know the distance between these 2 points is 4.24. However, I do not know how to find point C, which is 1/4 the distance from point A to point B. How do I find point C?

#2   Members   -  Reputation: 137

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Posted 17 April 2011 - 06:00 PM

What you are looking for is the midpoint - http://en.wikipedia.org/wiki/Midpoint

#3   Crossbones+   -  Reputation: 7526

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Posted 17 April 2011 - 06:09 PM

I am having a problem finding a point at a specific distance between 2 points. Suppose I have point A(1,1) and point B(4,4). I know the distance between these 2 points is 4.24. However, I do not know how to find point C, which is 1/4 the distance from point A to point B. How do I find point C?


to find a line 1/4th of the distance between 2 points you just need:

Vector2D p1(1,1);
Vector2D p2(4,4);

Vector2D direction = p2-p1

Vector2D p3 = p1+ direction* 0.25; (assuming your vector library supports scalar multiplication, if not its simply p3.x = p1.x + direction.x*0.25 and same for y)

If you don't have a vector library you can simply do

p1x=1;
p1y=1;
p2x=4;
p2y=4;

dirx=p2x-p1x;
diry=p2y-p1y;

p3x = p1x+dirx*0.25;
p3y = p1y+diry*0.25;
I don't suffer from insanity, I'm enjoying every minute of it.
The voices in my head may not be real, but they have some good ideas!

#4   Moderators   -  Reputation: 9438

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Posted 17 April 2011 - 06:11 PM

Using vector algebra:
C = (B - A) * .25 + A
C = (<4,4> - <1,1>) * .25 + <1,1>
C = <3,3> * .25 + <1,1>
C = <.75,.75> + <1,1>
C = <1.75, 1.75>

Or:
Cx = (Bx - Ax) * .25 + Ax
Cy = (By - Ay) * .25 + Ay
gzip: H4sIAAAAAAAEAG1QTUvEMBC991e8nvaiFfYoS7yo sLCo6MnjtJ1ugmkiyWRL/72z3T1YEQIJ8z4zA2Xp yPvt1qBpGrRFIJZkk9FyRyUzHCbKIHgn4hnZOrm1 TD0mG0HCCs+QGDGWziKXI6Wm2n++GYwUVH2mrGEE PnGCVQ8K8+JYfXA6URDEQfMZh5h6g5eoAlWJdeEI bbH2qYZf7XMUfw8f/Q0oMeZYNL9/WHF0uFEshvMr XYujd9SycFb+F18QcSOvlJauZ8ejqevdnV7/d550 e0t6prmunh73Bu+vz4c/XUeOQXfJgvKNkhf95U8/ Dtgmy5IBAAA=

#5   Members   -  Reputation: 103

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Posted 17 April 2011 - 06:43 PM

thank you very much!!




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