线段中点的方向

线段给定点 P 的方向只是意味着给定点 P 和线段（比如 AB）的坐标，我们必须从线段确定点 P 的方向。即点位于线段的右侧还是线段的左侧。

该点可能位于线段后面，在这种情况下，我们通过延长线段并确定点的方向来假设假想线。

* 只有三种情况，要么点在左侧，要么在右侧，要么在线段本身。
这是一个非常基本的问题，在线地图中的方向经常遇到，
示例：假设用户 A 必须去下图中的 C 点，该用户首先到达 B 点，但之后用户 A 如何知道他必须右转还是左转。

从线段知道点的方向也可以作为解决更复杂问题的基石，例如：

• 线段交点：判断两条线段是否相交
• 一组点的凸包。

我们将使用的坐标系是笛卡尔平面，因为大多数二维问题使用笛卡尔平面，因为这是一个二维问题。
这个问题可以用向量代数的叉积来解决
两点 A 和 B 的叉积为： A _x * B _y – A _y * B _x
其中A _x和A _y分别是A 的x 和y 坐标。类似地，B _x和 B _y分别是 B 的 x 和 y 坐标。
Cross-Product有一个有趣的属性，它将用于从线段确定点的方向。也就是说，当且仅当这些点在原点 (0, 0) 处的角度为逆时针方向时，两点的叉积为正。相反，当且仅当这些点在原点的角度为顺时针方向时，叉积为负。
一个例子肯定会澄清它，
在下图中，角度 BOP 为逆时针方向，BXP = 29*28 – 15*(-15) = 1037 的叉积为正。

这有助于我们得出结论，右侧的点必须具有正的叉积，而左侧的点必须具有负的叉积。还要注意，我们假设线段的一个点是原点，因此我们需要转换任何三点系统，使得线段的一个点是原点。

下面的例子解释了这个概念：
将A、B、P三点换算成A’、B’ 、 P’ ，以A为原点（可以简单地用P、B点减去A的坐标即可），然后计算交叉积：59*18 – (-25)*18 = 2187
由于这是正数，因此点 P 位于线段 AB 的右侧。

C++

// C++ Program to Determine Direction of Point
// from line segment
#include 
using namespace std;
 
// structure for point in cartesian plane.
struct point {
    int x, y;
};
 
// constant integers for directions
const int RIGHT = 1, LEFT = -1, ZERO = 0;
 
int directionOfPoint(point A, point B, point P)
{
    // subtracting co-ordinates of point A from
    // B and P, to make A as origin
    B.x -= A.x;
    B.y -= A.y;
    P.x -= A.x;
    P.y -= A.y;
 
    // Determining cross Product
    int cross_product = B.x * P.y - B.y * P.x;
 
    // return RIGHT if cross product is positive
    if (cross_product > 0)
        return RIGHT;
 
    // return LEFT if cross product is negative
    if (cross_product < 0)
        return LEFT;
 
    // return ZERO if cross product is zero.
    return ZERO;
}
 
// Driver code
int main()
{
    point A, B, P;
    A.x = -30;
    A.y = 10; // A(-30, 10)
    B.x = 29;
    B.y = -15; // B(29, -15)
    P.x = 15;
    P.y = 28; // P(15, 28)
 
    int direction = directionOfPoint(A, B, P);
    if (direction == 1)
        cout << "Right Direction" << endl;
    else if (direction == -1)
        cout << "Left Direction" << endl;
    else
        cout << "Point is on the Line" << endl;
    return 0;
}

Java

// Java Program to Determine Direction of Point
// from line segment
class GFG
{
 
// structure for point in cartesian plane.
static class point
{
    int x, y;
};
 
// constant integers for directions
static int RIGHT = 1, LEFT = -1, ZERO = 0;
 
static int directionOfPoint(point A,
                            point B, point P)
{
    // subtracting co-ordinates of point A
    // from B and P, to make A as origin
    B.x -= A.x;
    B.y -= A.y;
    P.x -= A.x;
    P.y -= A.y;
 
    // Determining cross Product
    int cross_product = B.x * P.y - B.y * P.x;
 
    // return RIGHT if cross product is positive
    if (cross_product > 0)
        return RIGHT;
 
    // return LEFT if cross product is negative
    if (cross_product < 0)
        return LEFT;
 
    // return ZERO if cross product is zero.
    return ZERO;
}
 
// Driver code
public static void main(String[] args)
{
    point A = new point(),
          B = new point(), P = new point();
    A.x = -30;
    A.y = 10; // A(-30, 10)
    B.x = 29;
    B.y = -15; // B(29, -15)
    P.x = 15;
    P.y = 28; // P(15, 28)
 
    int direction = directionOfPoint(A, B, P);
    if (direction == 1)
        System.out.println("Right Direction");
    else if (direction == -1)
        System.out.println("Left Direction");
    else
        System.out.println("Point is on the Line");
    }
}
 
// This code is contributed
// by Princi Singh

Python3

# Python3 program to determine direction
# of point from line segment
  
# Structure for point in cartesian plane.
class point:
     
    def __init__(self):
         
        self.x = 0
        self.y = 0
  
# Constant integers for directions
RIGHT = 1
LEFT = -1
ZERO = 0
  
def directionOfPoint(A, B, P):
     
    global RIGHT, LEFT, ZERO
     
    # Subtracting co-ordinates of
    # point A from B and P, to
    # make A as origin
    B.x -= A.x
    B.y -= A.y
    P.x -= A.x
    P.y -= A.y
  
    # Determining cross Product
    cross_product = B.x * P.y - B.y * P.x
  
    # Return RIGHT if cross product is positive
    if (cross_product > 0):
        return RIGHT
         
    # Return LEFT if cross product is negative
    if (cross_product < 0):
        return LEFT
  
    # Return ZERO if cross product is zero
    return ZERO
 
# Driver code
if __name__=="__main__":
     
    A = point()
    B = point()
    P = point()
     
    A.x = -30
    A.y = 10 # A(-30, 10)
    B.x = 29
    B.y = -15 # B(29, -15)
    P.x = 15
    P.y = 28 # P(15, 28)
  
    direction = directionOfPoint(A, B, P)
     
    if (direction == 1):
        print("Right Direction")
    elif (direction == -1):
        print("Left Direction")
    else:
        print("Point is on the Line")
 
# This code is contributed by rutvik_56

C#

// C# Program to Determine Direction of Point
// from line segment
using System;
using System.Collections.Generic;
 
class GFG
{
 
// structure for point in cartesian plane.
public class point
{
    public int x, y;
};
 
// constant integers for directions
static int RIGHT = 1, LEFT = -1, ZERO = 0;
 
static int directionOfPoint(point A,
                            point B, point P)
{
    // subtracting co-ordinates of point A
    // from B and P, to make A as origin
    B.x -= A.x;
    B.y -= A.y;
    P.x -= A.x;
    P.y -= A.y;
 
    // Determining cross Product
    int cross_product = B.x * P.y - B.y * P.x;
 
    // return RIGHT if cross product is positive
    if (cross_product > 0)
        return RIGHT;
 
    // return LEFT if cross product is negative
    if (cross_product < 0)
        return LEFT;
 
    // return ZERO if cross product is zero.
    return ZERO;
}
 
// Driver code
public static void Main(String[] args)
{
    point A = new point(),
          B = new point(),
          P = new point();
    A.x = -30;
    A.y = 10; // A(-30, 10)
    B.x = 29;
    B.y = -15; // B(29, -15)
    P.x = 15;
    P.y = 28; // P(15, 28)
 
    int direction = directionOfPoint(A, B, P);
    if (direction == 1)
        Console.WriteLine("Right Direction");
    else if (direction == -1)
        Console.WriteLine("Left Direction");
    else
        Console.WriteLine("Point is on the Line");
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// JavaScript Program to Determine Direction of Point
// from line segment
 
// structure for point in cartesian plane.
class point
{
    constructor()
    {
        this.x=0;
        this.y=0;
    }
}
 
// constant integers for directions
let RIGHT = 1, LEFT = -1, ZERO = 0;
 
function directionOfPoint(A,B,P)
{
    // subtracting co-ordinates of point A
    // from B and P, to make A as origin
    B.x -= A.x;
    B.y -= A.y;
    P.x -= A.x;
    P.y -= A.y;
  
    // Determining cross Product
    let cross_product = B.x * P.y - B.y * P.x;
  
    // return RIGHT if cross product is positive
    if (cross_product > 0)
        return RIGHT;
  
    // return LEFT if cross product is negative
    if (cross_product < 0)
        return LEFT;
  
    // return ZERO if cross product is zero.
    return ZERO;
}
 
// Driver code
let A = new point(),
          B = new point(), P = new point();
    A.x = -30;
    A.y = 10; // A(-30, 10)
    B.x = 29;
    B.y = -15; // B(29, -15)
    P.x = 15;
    P.y = 28; // P(15, 28)
  
    let direction = directionOfPoint(A, B, P);
    if (direction == 1)
        document.write("Right Direction<br>");
    else if (direction == -1)
        document.write("Left Direction");
    else
        document.write("Point is on the Line");
 
 
// This code is contributed by rag2127
 

输出：

Right Direction

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