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Index › Programming Questions & Help › Syntax Questions › direction of a line; mid point; perpindicular?
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direction of a line; mid point; perpindicular??? (Read 354 times)
jce
direction of a line; mid point; perpindicular???
Aug 10th, 2006, 3:22am  
sorry, im completely lost on this one and was wondering if someone here wouldn't mind helping me out w/ a few pointers...

basically, if im given a line, i'd like to know how to calculate what the mid point on the line is, and which direction its pointing so that i can create another line that intersects it perpendicularly.

Quote:

void setup(){
size (200,200);
line (40,100,160,100);//this one is easy to find the middle point,and create a line that intersects it at the middle perpendicularly
stroke (255,0,0);
line (100,75,100,125);
stroke (255,0,255);
line (14,65,80,10);//but this one? no idea how to find mid point, or to make a line intersect it perpendicularly.
}

void draw(){
}


i know it would be able to draw a cross, then transform it by rotating, but thats not what i need for this application.
any help anyone could give would be much appreciated

cheers!

jc
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JohnG
Re: direction of a line; mid point; perpindicular?
Reply #1 - Aug 10th, 2006, 11:30am  
Here's some code that should hopefully help:

Code:
float startX,startY,endX,endY;
startX=10;
startY=10;
endX=190;
endY=190;

midX=startX+((endX-startX)/2.0);
midY=startY+((endY-startY)/2.0);

float dir=atan2(endY-startY,endX-startX);
float rotatedDir=dir+PI/2.0;

float lineLength=sqrt(sq(endX-startX)+sq(endY-startY));

float newStartX=midX+( (lineLength/2.0) * cos(rotatedDir) );
float newStartY=midY+( (lineLength/2.0) * sin(rotatedDir) );
float newEndX=midX-( (lineLength/2.0) * cos(rotatedDir) );
float newEndY=midY-( (lineLength/2.0) * sin(rotatedDir) );


The above was all just typed and not tested, so may have a few small errors, but I think should work and illustrates the actions needed to be performed to do what you want.
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jce
Re: direction of a line; mid point; perpindicular?
Reply #2 - Aug 10th, 2006, 7:33pm  
brilliant dude thats exactly what i was looking for.
thank you very much!

-jc
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