问题
In Java, I have a class Line
that has two variables : m
and b
, such that the line follows the formula mx + b
. I have two such lines. How am I to find the x
and y
coordinates of the intersection of the two lines? (Assuming the slopes are different)
Here is 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;
}
}
回答1:
Lets assume you have these 2 functions:
y = m1*x + b1
y = m2*x + b2
To find the intersection point of the x-axis
we do:
m1*x+b1 = m2*x+b2
m1*x-m2*x = b2 - b2
x(m1-m2) = (b2-b1)
x = (b2-b1) / (m1-m2)
To find y, you use of the function expressions and replace x
for its value (b2-b1) / (m1-m2)
.
So:
y = m1 * [(b2-b1) / (m1-m2)] + b1
You have (this.b - line.b)
, change to (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);
}
回答2:
That's what i got. Couldn't find any exeptions which don't work:
public static Point calculateInterceptionPoint(Point s1, Point d1, Point s2, Point d2) {
double sNumerator = s1.y * d1.x + s2.x * d1.y - s1.x * d1.y - s2.y * d1.x;
double sDenominator = d2.y * d1.x - d2.x * d1.y;
// parallel ... 0 or infinite points, or one of the vectors is 0|0
if (sDenominator == 0) {
return null;
}
double s = sNumerator / sDenominator;
double t;
if (d1.x != 0) {
t = (s2.x + s * d2.x - s1.x) / d1.x;
} else {
t = (s2.y + s * d2.y - s1.y) / d1.y;
}
Point i1 = new Point(s1.x + t * d1.x, s1.y + t * d1.y);
return i1;
}
public static void main(String[] args) {
System.out.println(calculateInterceptionPoint(new Point(3, 5), new Point(0, 2), new Point(1, 2), new Point(4, 0)));
System.out.println(calculateInterceptionPoint(new Point(3, 5), new Point(0, 2), new Point(1, 2), new Point(0, 2)));
System.out.println(calculateInterceptionPoint(new Point(0, 0), new Point(0, 2), new Point(0, 0), new Point(2, 0)));
System.out.println(calculateInterceptionPoint(new Point(0, 0), new Point(0, 2), new Point(0, 0), new Point(0, 2)));
System.out.println(calculateInterceptionPoint(new Point(0, 0), new Point(0, 0), new Point(0, 0), new Point(0, 0)));
}
回答3:
The proposed solution by @wutzebaer seems not to work, instead try the solution below (code based on the example from: https://rosettacode.org/wiki/Find_the_intersection_of_two_lines#Java). s1 and s2 are the endpoints of the first line and d1 and d2 are the endpoints of the second line.
public static Point2D.Float calculateInterceptionPoint(Point2D.Float s1, Point2D.Float s2, Point2D.Float d1, Point2D.Float d2) {
double a1 = s2.y - s1.y;
double b1 = s1.x - s2.x;
double c1 = a1 * s1.x + b1 * s1.y;
double a2 = d2.y - d1.y;
double b2 = d1.x - d2.x;
double c2 = a2 * d1.x + b2 * d1.y;
double delta = a1 * b2 - a2 * b1;
return new Point2D.Float((float) ((b2 * c1 - b1 * c2) / delta), (float) ((a1 * c2 - a2 * c1) / delta));
}
public static void main(String[] args) {
System.out.println(calculateInterceptionPoint(new Point2D.Float(3, 5), new Point2D.Float(0, 2), new Point2D.Float(1, 2), new Point2D.Float(4, 0)));
}
来源:https://stackoverflow.com/questions/31506740/java-find-intersection-of-two-lines