「java直线斜率」垂直的线 斜率
今天给各位分享java直线斜率的知识，其中也会对垂直的线 斜率进行解释，如果能碰巧解决你现在面临的问题，别忘了关注本站，现在开始吧！
本文目录一览：
- 1、用JAVA求一条直线的斜率
- 2、java 判断直线状况
- 3、java求高手改错
- 4、java编程判断这两条直线是否相交？，若相交，求出交点。
- 5、用JAVA语言证明两条直线平行？
- 6、JAVA直线求斜率的时候为什么要求公约数
用JAVA求一条直线的斜率
k
=
tan
a
//斜率
tan
a
=
（y2-y1）/（x2-x1）//a为度数，(x1,y1),(x2,y2)分别为直线2个点的坐标
public
class
Test
{
public
Double
k;
//斜率
//计算斜率
public
Double
countK(Double
x1,Double
y1,Double
x2,Double
y2){
return
(y2-y1)/(x2-x1);
}
public
static
void
main(String[]
args)
{
System.out.println(new
Test().countK(3.0,
4.0,
5.0,
6.0));
}
}
java 判断直线状况
public class LineTest {
public static void main(String[] args) {
Point p1 = new Point(0, 3), p2 = new Point(4, 0);
Line l1 = new Line(0, 3, 4, 0),
l2 = new Line(p1, p2),
l3 = new Line(0,4,0,3),
l4 = new Line(4,4,0,4),
l5 = new Line(0,0,3,3);
System.out.println("l1:"+l1);
System.out.println("l2:"+l2);
System.out.println("l3:"+l3);
System.out.println("l4:"+l4);
System.out.println("l5:"+l5);
System.out.println("l1 length:"+l1.getLength());
System.out.println("l2 length:"+l2.getLength());
System.out.println("l3 length:"+l3.getLength());
System.out.println("l4 length:"+l4.getLength());
System.out.println("l1 isHorizontal:"+l1.isHorizontal());
System.out.println("l3 isHorizontal:"+l3.isHorizontal());
System.out.println("l1 isVertical:"+l1.isVertical());
System.out.println("l4 isVertical:"+l4.isVertical());
System.out.println("l1 slope:"+l1.getSlope());
System.out.println("l3 slope:"+l3.getSlope());
System.out.println("l4 slope:"+l4.getSlope());
System.out.println("l5 slope:"+l5.getSlope());
System.out.println("l1 midpoint:"+l1.getMidpoint().toString());
System.out.println("l3 midpoint:"+l3.getMidpoint().toString());
System.out.println("l1 equals l2:"+l1.equals(l2));
System.out.println("l1 equals l3:"+l1.equals(l3));
System.out.println("l2 equals l3:"+l2.equals(l3));
}
}
class Line {
private Point p1,p2;
public Line(Point p1, Point p2) {
super();
this.p1 = p1;
this.p2 = p2;
}
public Line(int x1,int y1,int x2,int y2){
super();
this.p1=new Point(x1, y1);
this.p2=new Point(x2, y2);
}
public int getLength(){
int a=(int) Math.abs(p1.getX()-p2.getX());
int b=(int) Math.abs(p1.getY()-p2.getY());
return (int) Math.sqrt(a*a+b*b);
}
public boolean isHorizontal(){
return Math.abs(p1.getX()-p2.getX())=0.00001;
}
public boolean isVertical(){
return Math.abs(p1.getY()-p2.getY())=0.00001;
}
public double getSlope(){
double a=p1.getX()-p2.getX();
double b=p1.getY()-p2.getY();
if(b==0.0)return Double.MAX_VALUE;
return a/b;
}
public Point getMidpoint(){
return new Point((int)((p1.getX()+p2.getX())/2)
,(int)((p1.getY()+p2.getY())/2));
}
public boolean equals(Line ol){
return (this.p1.equals(ol.getP1()))
(this.p2.equals(ol.getP2()));
}
public Point getP1() {
return p1;
}
public void setP1(Point p1) {
this.p1 = p1;
}
public Point getP2() {
return p2;
}
public void setP2(Point p2) {
this.p2 = p2;
}
@Override
public String toString(){
StringBuilder sb=new StringBuilder();
sb.append("[")
.append("(").append(p1.getX()).append(",")
.append(p1.getY()).append(")").append(",")
.append("(").append(p2.getX()).append(",")
.append(p2.getY()).append(")")
.append("]");
return sb.toString();
}
}
有个小问题就是你要求用Point做点，而Point点的数据类型是int，计算中点时会损失精度，可以改成Point2D实现。
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
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：
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直线求斜率的时候为什么要求公约数
因为例如up=4，down=2，和up=8，down=4,计算出来的斜率是一样的，这俩种情况的取值可能会出现在一条直线上。返回值则是a和b的最大公约数，所以JAVA直线求斜率的时候要求公约数。
java直线斜率的介绍就聊到这里吧，感谢你花时间阅读本站内容，更多关于垂直的线 斜率、java直线斜率的信息别忘了在本站进行查找喔。
发布于：2022-11-28，除非注明，否则均为原创文章，转载请注明出处。