重庆分公司,新征程启航

为企业提供网站建设、域名注册、服务器等服务

java判断平行代码 java或判断

java 判断一个平面两条直线是否平行

两点成一线,两线成一面

站在用户的角度思考问题,与客户深入沟通,找到平泉网站设计与平泉网站推广的解决方案,凭借多年的经验,让设计与互联网技术结合,创造个性化、用户体验好的作品,建站类型包括:网站设计、成都做网站、企业官网、英文网站、手机端网站、网站推广、域名与空间、网站空间、企业邮箱。业务覆盖平泉地区。

如果在坐标系里面,可以根据坐标判断啊

java编程判断这两条直线是否相交?,若相交,求出交点。

您好,下面是我写的代码,麻烦看一下是否符合要求。

Point类:

public class Point {

private int x;

private int y;

public int getX() {

  return x;

}

public void setX(int x) {

  this.x = x;

}

public int getY() {

  return y;

}

public void setY(int y) {

  this.y = y;

}

}

Line类:

public class Line {

private Point a;

private Point b;

public Point getA() {

  return a;

}

public Point getB() {

  return b;

}

public Line(Point a, Point b) {

  this.a = a;

  this.b = b;

}

}

LineIntersect类:

public class LineIntersect {

private Line ab;

private Line cd;

public LineIntersect(Line ab, Line cd) {

  this.ab = ab;

  this.cd = cd;

}

public void JudgeLineIntersect() {

  Point A = this.ab.getA();

  Point B = this.ab.getB();

  Point C = this.cd.getA();

  Point D = this.cd.getB();

  int d1 = (B.getY() - A.getY()) * (D.getX() - C.getX());

  int d2 = (B.getX() - A.getX()) * (D.getY() - C.getY());

  int d3 = (C.getY() - A.getY()) * (B.getX() - A.getX());

  int d4 = (C.getX() - A.getX()) * (B.getY() - A.getY());

  if (d1 == d2) {

      System.out.println("两条直线平行。");

  } else if (d3 == d4) {

      System.out.println("两条直线重合。");

  } else {

      System.out.println("两条直线相交。");

      int d5 = C.getY() * D.getX() - C.getX() * D.getY();

      int d6 = A.getY() * B.getX() - A.getX() * B.getY();

      float intersect_x = ((C.getX() - D.getX()) * d6 - (A.getX() - B.getX()) * d5)

              / ((B.getY() - A.getY()) * (D.getX() - C.getX()) - (B.getX() - A.getX()) * (D.getY() - C.getY()));

      float intersect_y = ((C.getY() - D.getY()) * d6 - (A.getY() - B.getY()) * d5)

              / ((B.getY() - A.getY()) * (D.getX() - C.getX()) - (B.getX() - A.getX()) * (D.getY() - C.getY()));

      System.out.println("交点的X坐标为:" + intersect_x);

      System.out.println("交点的Y坐标为:" + intersect_y);

  }

}

}

下面是主类(测试类):

import java.util.Scanner;

public class Main {

public static void main(String[] args) {

  System.out.print("请输入A点的X坐标:");

  Scanner inAX = new Scanner(System.in);

  int ax = inAX.nextInt();

  System.out.print("请输入A点的Y坐标:");

  Scanner inAY = new Scanner(System.in);

  int ay = inAY.nextInt();

  System.out.print("请输入B点的X坐标:");

  Scanner inBX = new Scanner(System.in);

  int bx = inBX.nextInt();

  System.out.print("请输入B点的Y坐标:");

  Scanner inBY = new Scanner(System.in);

  int by = inBY.nextInt();

  System.out.print("请输入C点的X坐标:");

  Scanner inCX = new Scanner(System.in);

  int cx = inCX.nextInt();

  System.out.print("请输入C点的Y坐标:");

  Scanner inCY = new Scanner(System.in);

  int cy = inCY.nextInt();

  System.out.print("请输入D点的X坐标:");

  Scanner inDX = new Scanner(System.in);

  int dx = inDX.nextInt();

  System.out.print("请输入D点的Y坐标:");

  Scanner inDY = new Scanner(System.in);

  int dy = inDY.nextInt();

  Point A = new Point();

  A.setX(ax);

  A.setY(ay);

  Point B = new Point();

  B.setX(bx);

  B.setY(by);

  Point C = new Point();

  C.setX(cx);

  C.setY(cy);

  Point D = new Point();

  D.setX(dx);

  D.setY(dy);

  Line AB = new Line(A, B);

  Line CD = new Line(C, D);

  LineIntersect lineIntersect = new LineIntersect(AB, CD);

  lineIntersect.JudgeLineIntersect();

  inAX.close();

  inAY.close();

  inBX.close();

  inBY.close();

  inCX.close();

  inCY.close();

  inDX.close();

  inDY.close();

}

}

下面是我刚才做的测试结果:

java输入四个坐标判断是否为平行四边形并显示该图形.QQ847507674会的大侠教下急用

找下平行四边形判定条件,根据条件判定就行了啊!至于作图,你在判断的过程中就能知道哪些顶点应该相连,在该相连的顶点间作直线就可以了。

用JAVA语言证明两条直线平行?

两直线平行的条件:

如果两条直线的斜率都存在, 并且相等, 那么这两条直线平行. (y=2x + 1 and y=2x - 1)

如果两条直线的斜率都不存在, 那么这两条直线平行. (x = 1 and x= -1)

以上有2种特殊情况需要注意:

a. 斜率都为0, 例如 y= 1 and y= -1.

b. 斜率不存在, 例如 x=1 and x= -1.

这时, 两直线也都平行.

那么用java 实现, 我们就需要用户以y=kx + b, 中的k值, 即斜率.

import java.util.Scanner;

public class ParallelismDemo {

public static void main(String args[])

{

new ParallelismDemo();

}

//Constructor method, will run during newing instance of this class. 

public ParallelismDemo()

{

Scanner scan = new Scanner(System.in);

System.out.println("Please enter first line's value 'k' which format is y = kx + b (If not exist, enter n)");

String firstValueK = scan.nextLine();

System.out.println("Please enter second line's value 'k' which format is y = kx + b (If not exist, enter n)");

String secValueK = scan.nextLine();

scan.close();

//Check both k are not exist

if(firstValueK.equals("n")  secValueK.equals("n"))

{

System.out.println("Your lines are parallel");

return;

}

// If both k exist, then check whether they are equals

else if(firstValueK.equals(secValueK))

{

System.out.println("Your lines are parallel");

return;

}

// For the rest condition, they are not parallel

else 

{

System.out.println("Your lines are not parallel");

return;

}

}

}

java求高手改错

其实如一楼所说这种最好还是用2个点定义一条直线必要好点。但是根据题目的要求。我完善和更改了一些漏洞。都写了注释的。数学不是很好,可能还有一些地方没有想到。代码如下:

=====================================================

public class Line {

private Point point;

private double slope;

public Line(Point point, double slope) {

this.setPoint(point);

this.setSlope(slope);

}

public Line(Point p1, Point p2) {

this(p1, 0);

if(p1.getX() == p2.getX()){ // 判断是否斜率存在。不存在斜率设置为无穷大 否则设置为计算所得斜率

this.setSlope(Double.MAX_VALUE);

}

else{

this.setSlope((double)(p2.getY() - p1.getY())/(double)(p2.getX() - p1.getX()));

}

}

public Line(int a, int b) {

this(new Point(a, 0), new Point(0, b));

}

public void setSlope(double slope) {

this.slope = slope;

}

public void setPoint(Point point) {

this.point = point;

}

public double getSlope() {

return slope;

}

public Point getPoint() {

return point;

}

public boolean isParallel(Line line) {

if(this.point.isSamePoint(line.point)){ // 同一条直线重合 或者已经相交 故不平行

return false;

}

else{

if(this.isPointOnLine(line.point)){ // 如果两条直线的定义点都在各自的线上那么说明他们重合不平行

return false;

}

else{

return this.slope == line.slope; //否则才判断斜率是否相等

}

}

}

public boolean isPointOnLine(Point p){

double tmp = (double)(this.point.y - p.getY())/(double)(this.point.x - p.getX());

if(this.slope == tmp){

return true;

}

return false;

}

public static void main(String[] args) {

Line l1 = new Line(new Point(1, 0), new Point(1, 3));

Line l2 = new Line(new Point(2, 0), new Point(2, 3));

System.out.println(l1.isParallel(l2));

}

}

class Point{

int x,y;

public Point(int a,int b){

this.x = a;

this.y = b;

}

public int getX(){

return x;

}

public int getY(){

return y;

}

public boolean isSamePoint(Point p){

if(this.x == p.getX() this.y == p.getY()){

return true;

}

return false;

}

}

=============================

程序执行结果为 true


新闻标题:java判断平行代码 java或判断
网站网址:http://cqcxhl.cn/article/hhssci.html

其他资讯

在线咨询
服务热线
服务热线:028-86922220
TOP