Calculating y-intercept of a straight line

Programming Questions

petrus250

Calculating y-intercept of a straight line

in Programming Questions • 1 year ago

Hi,

I need to calculate the y intercept of a straight line given 2 sets of coordinates and the gradient, although this obviously has to be done automatically... here is my current code:

void getEquation(float gradient) {
float intercept = ;
String output = gradient + "x + " + intercept;
println(output);
}

Replies(3)

PhiLho

Re: Calculating y-intercept of a straight line

1 year ago

Where are the two sets of coordinates in your code? What the gradient is doing in there? What do you expect the intercept variable to be?

Perhaps a little drawing might explain this better.

petrus250

Re: Re: Calculating y-intercept of a straight line

1 year ago

without the gradient I would need some much more advanced maths to calculate the y-intercept, and the 2 sets of coords are pointX and pointY for x1 and y1 and the x2 and y2 are mouseX and mouseY (it is dynamically calculated as the mouse is moved.

benja

Re: Calculating y-intercept of a straight line

1 year ago

You can easily calculate the gradient if you have two points: gradient = (y2 - y1) / (x2 - x1)

Here is a simple example which might help:

PVector p1 = new PVector(150, 30);
PVector p2 = new PVector(30, 70);
PVector p3 = new PVector();

float a, b;

void setup() {
size(300, 300);
background(0);
fill(255, 0, 0);
stroke(255);

getEquation();
p3 = getYIntercept();
}

void draw() {
line(p2.x, p2.y, p3.x, p3.y);
ellipse(p1.x, p1.y, 5, 5);
ellipse(p2.x, p2.y, 5, 5);
ellipse(p3.x, p3.y, 5, 5);
}

void mousePressed() {
background(0);
p2.x = mouseX;
p2.y = mouseY;
getEquation();
p3 = getYIntercept();
}

// calculate equation (y = a * x +b)
void getEquation() {
a = (p2.y-p1.y)/(p2.x-p1.x);
b = p1.y - a *p1.x;
println("gradient: "+a);
println("intercept: "+b);
}

// calculate point on line for y = 0
PVector getYIntercept() {
float x = -b /a;
return new PVector(x, 0);
}

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