Java查找兩條線的交點

Question

在Java中，我有一個類Line，它有兩個變量：m和b，使得該行遵循公式mx + b。我有兩條這樣的路線。我如何找到兩條線的交點座標爲x和y？ （假設斜率不同）

這裏是class Line：

import java.awt.Graphics; 
import java.awt.Point; 

public final class Line { 
    public final double m, b; 

    public Line(double m, double b) { 
     this.m = m; 
     this.b = b; 
    } 

    public Point intersect(Line line) { 
     double x = (this.b - line.b)/(this.m - line.m); 
     double y = this.m * x + this.b; 
     return new Point((int) x, (int) y); 
    } 

    public void paint(Graphics g, int startx, int endx, int width, int height) { 
     startx -= width/2; 
     endx -= width/2; 
     int starty = this.get(startx); 
     int endy = this.get(endx); 
     Point points = Format.format(new Point(startx, starty), width, height); 
     Point pointe = Format.format(new Point(endx, endy), width, height); 
     g.drawLine(points.x, points.y, pointe.x, pointe.y); 
    } 

    public int get(int x) { 
     return (int) (this.m * x + this.b); 
    } 

    public double get(double x) { 
     return this.m * x + this.b; 
    } 
} 

Answer 1

讓我們假設你有這兩個功能：

y = m1*x + b1  
y = m2*x + b2 

要找到x-axis的交匯點上，我們做的事：

m1*x+b1 = m2*x+b2  
m1*x-m2*x = b2 - b2  
x(m1-m2) = (b2-b1)  
x = (b2-b1)/(m1-m2) 

要找到y，可以使用函數exp並將x替換爲其值(b2-b1)/(m1-m2)。

所以：

y = m1 * [(b2-b1)/(m1-m2)] + b1 

你有(this.b - line.b)，改變(line.b - this.b)。

public Point intersect(Line line) { 
    double x = (line.b - this.b)/(this.m - line.m); 
    double y = this.m * x + this.b; 

    return new Point((int) x, (int) y); 
}