Java找到两条线的交集

Question

在Java中,我有一个Line具有两个变量的类:m并且b,这样的行遵循公式mx + b.我有两条这样的台词.我如何找到两条线的交点x和y坐标？(假设斜坡不同)

这是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

让我们假设你有这两个功能:

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

要找到x-axis我们所做的交点:

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

要查找y,可以使用函数表达式并替换x其值(b2-b1) / (m1-m2).

所以:

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

你有(this.b - line.b),改为(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);
}