给定点P从线段的方向仅表示给定点P和线段的坐标(例如AB)，我们必须确定线段P的方向。那就是该点位于线段的右侧还是线段的左侧。
该点可能位于线段的后面，在这种情况下，我们通过延伸线段并确定点的方向来假设一条假想线。
*只有三种情况，该点在左侧，右侧或在线段本身上。
这是一个非常基本的问题，在线地图中的路线通常会遇到该问题，
示例：假设用户A必须转到下图中的点C，该用户首先到达点B，但此后用户A如何知道他必须向右转还是向左转。
从线段知道点的方向还可以作为构建模块来解决更复杂的问题，例如：
- 线段相交：查找两个线段是否相交
- 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都使用笛卡尔平面，并且由于这是二维问题。
向量代数的叉积可以解决此问题
两个点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。同样，B x和B y分别是B的x和y坐标。
叉积具有有趣的属性，可用于确定线段中点的方向。也就是说，当且仅当这些点在原点(0，0)处的角度为逆时针时，两个点的叉积才为正。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这一点，
在下图中，角度BOP为逆时针，并且BXP的叉积= 29 * 28 – 15 *(-15)= 1037，为正。
这有助于我们得出以下结论：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为原点，因此我们需要转换任何三点系统，以使线段的一个点为原点。
以下示例解释了该概念：
将三个点A，B和P转换为A’，B’和P’ ，以A为原点(这可以简单地通过从点P和B减去A的坐标来完成)，然后计算叉积：59 * 18 –(-25)* 18 = 2187
由于这是肯定的，所以点P在线段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 SinghPython3
# 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_56C#
// 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输出：
Right Direction