Line help beginner completely new

Processing Coding Questions
1

How to make a line that is at the same position as mouse but is longer than the distance. belowimg

1 Like
2

Maye this helps?

float x, y;

void setup () {
  size (400, 400);
  background (255);
  x = width/2;
  y = height/2;
}

void draw () {
  background (255);
  line (x, y, mouseX, mouseY);
}
3

Sorry, but didn’t help. I need it to be farther than the mouse postion.

4

But from which starting point?

5

I have an ellipse(x, y, 20, 20) and from the ellipse’s position I want it to start and I want not to end the line, like on the image, where is the line longer than the distance between mouse and ellipse.

6

Subtracting or adding from mouseX and mouseY with some value?

7

And I want the line to be pointing at my cursor and not to end the line when it hits the mouse, but continue like on the image.

8

Hehe, for me sometimes to understand the question is harder than the code.

9

What do you mean by this. Normally the ponter points. Not the line.

10

I mean like it will be where my cursor is, look at the image.

Also, I want the line to not end but has infinite distance. Like it will never end, I don’t want it to be long from the ellipse to my Cursor but farther than my cusror

11

But I already replied just add a value to MouseX and MouseY.

12

Doesn’t work.

But if you can’t understand what I am asking for, there is a video where they have the neverending line. I’m gonna link it here > https://youtu.be/v1kqMxsvElM?t=265

1 Like
13

This should do what you want.

int px, py; // ellipse position
float len;
void setup(){
  size(640, 480);
  cursor(CROSS);
  len = max(width, height);
  px = 255;
  py = 199;
}

void draw(){
  background(255, 240, 200);
  drawLine(px, py, mouseX, mouseY);
  fill(0);
  noStroke();
  ellipse(px, py, 10, 10);
}

// Draw a line from [fx, fy] to pass through [tx, ty]
// to the display edge
void drawLine(float fx, float fy, float tX, float tY){
  float ang = atan2(tY - fy, tX - fx);
  float x = fx + cos(ang) * len;
  float y = fy + sin(ang) * len;
  stroke(255,0,0);
  strokeWeight(3);
  line(fx, fy, x, y);
}
3 Likes
14

Just to restate for future searches: OP wanted to:

  • draw a line from an origin THROUGH the mouse and PAST the edge of the display.

@quark’s solution mostly works, so long as px/py isn’t in a corner – like (10,10). Then it fails on the diagonal. For a line that is longer than the box, use:

len = sqrt(width*width + height*height)+1;

If you want to measure a line segment to exactly the edge of a bounding box that contains its endpoint – for example, if there is a rectangle in the middle of your screen, and you want to crop the line exactly – then you can use line-line collision detection to find its other endpoint.

// LINE/LINE
PVector lineLine(float x1, float y1, float x2, float y2, float x3, float y3, float x4, float y4) {

  // calculate the direction of the lines
  float uA = ((x4-x3)*(y1-y3) - (y4-y3)*(x1-x3)) / ((y4-y3)*(x2-x1) - (x4-x3)*(y2-y1));
  float uB = ((x2-x1)*(y1-y3) - (y2-y1)*(x1-x3)) / ((y4-y3)*(x2-x1) - (x4-x3)*(y2-y1));

  // if uA and uB are between 0-1, lines are colliding
  if (uA >= 0 && uA <= 1 && uB >= 0 && uB <= 1) {
    float ix = x1 + (uA * (x2-x1));
    float iy = y1 + (uA * (y2-y1));
    return new PVector(ix, iy);
  }
  return null;
}
15

continued here:

1 Like
16

Thanks for pointing that out I had not checked the corner position, obviously len is too small but calculating the diagonal of the screen is more than necessary simply change line to

len = 1.5 * max(width, height);

this will work for any size display. :slight_smile:

(this can be proved using basic algebra)

1 Like