在给定斜率的线上找到给定距离处的点

给定二维点 p(x ₀ , y ₀ ) 的坐标。找出离它距离为 L 的点，使得连接这些点形成的线的斜率为 M。

例子：

Input : p = (2, 1)
        L = sqrt(2)
        M = 1
Output :3, 2
        1, 0
Explanation:
The two points are sqrt(2) distance away 
from the source and have the required slope
m = 1.

Input : p = (1, 0)
        L = 5
        M = 0
Output : 6, 0
        -4, 0

我们需要在斜率为 M 的直线上找到与给定点相距 L 的两个点。
这个想法已在下面的帖子中介绍。
使用中点找到矩形的角

根据输入斜率，问题可以分为 3 类。

如果斜率为零，我们只需要调整源点的x坐标
如果斜率无限大，我们需要调整y坐标
对于斜率的其他值，我们可以使用以下等式来找到点

$Given \ that \ the \ point (x, y) \ is \ at \ distance \ I \ away \ from \ (x_0, y_0) \newline \newline (y-y_0)^{2} + (x-x_0)^{2}= l^{2} \newline \newline Also \ as \ the \ line \ that \ passes \ through \ (x, y) \ and \ (x0, y0) \ satisfies \newline \newline \frac{y-y_0}{x-x_0}= m \newline \newline Rearranging \ we \ get \newline y=y_0+m*(x-x_0) \newline \newline Putting \ the \ values \ in \ first \ equation \newline \newline m^2.(x-x_0)^2+(x-x_0)^2=l^2 \newline \newline Hence, \ we \ have \newline \newline x=x_0\pm l.\sqrt{\frac{1}{1+m^2}} \newline \newline y=y_0 \pm m.l.\sqrt{\frac{1}{1+m^2}}$

现在使用上面的公式我们可以找到所需的点。

C++

// C++ program to find the points on a line of
// slope M at distance L
#include 
using namespace std;
 
// structure to represent a co-ordinate
// point
struct Point {
 
    float x, y;
    Point()
    {
        x = y = 0;
    }
    Point(float a, float b)
    {
        x = a, y = b;
    }
};
 
// Function to print pair of points at
// distance 'l' and having a slope 'm'
// from the source
void printPoints(Point source, float l,
                                 int m)
{
    // m is the slope of line, and the
    // required Point lies distance l
    // away from the source Point
    Point a, b;
 
    // slope is 0
    if (m == 0) {
        a.x = source.x + l;
        a.y = source.y;
 
        b.x = source.x - l;
        b.y = source.y;
    }
 
    // if slope is infinite
    else if (m == std::numeric_limits
                                 ::max()) {
        a.x = source.x;
        a.y = source.y + l;
 
        b.x = source.x;
        b.y = source.y - l;
    }
    else {
        float dx = (l / sqrt(1 + (m * m)));
        float dy = m * dx;
        a.x = source.x + dx;
        a.y = source.y + dy;
        b.x = source.x - dx;
        b.y = source.y - dy;
    }
 
    // print the first Point
    cout << a.x << ", " << a.y << endl;
 
    // print the second Point
    cout << b.x << ", " << b.y << endl;
}
 
// driver function
int main()
{
    Point p(2, 1), q(1, 0);
    printPoints(p, sqrt(2), 1);
    cout << endl;
    printPoints(q, 5, 0);
    return 0;
}

Java

// Java program to find the points on 
// a line of slope M at distance L
class GFG{
 
// Class to represent a co-ordinate
// point
static class Point
{
    float x, y;
    Point()
    {
        x = y = 0;
    }
    Point(float a, float b)
    {
        x = a;
        y = b;
    }
};
 
// Function to print pair of points at
// distance 'l' and having a slope 'm'
// from the source
static void printPoints(Point source,
                        float l, int m)
{
     
    // m is the slope of line, and the
    // required Point lies distance l
    // away from the source Point
    Point a = new Point();
    Point b = new Point();
     
    // Slope is 0
    if (m == 0)
    {
        a.x = source.x + l;
        a.y = source.y;
 
        b.x = source.x - l;
        b.y = source.y;
    }
 
    // If slope is infinite
    else if (Double.isInfinite(m))
    {
        a.x = source.x;
        a.y = source.y + l;
 
        b.x = source.x;
        b.y = source.y - l;
    }
    else
    {
        float dx = (float)(l / Math.sqrt(1 + (m * m)));
        float dy = m * dx;
        a.x = source.x + dx;
        a.y = source.y + dy;
        b.x = source.x - dx;
        b.y = source.y - dy;
    }
 
    // Print the first Point
    System.out.println(a.x + ", " + a.y);
 
    // Print the second Point
    System.out.println(b.x + ", " + b.y);
}
 
// Driver code
public static void main(String[] args)
{
    Point p = new Point(2, 1),
          q = new Point(1, 0);
    printPoints(p, (float)Math.sqrt(2), 1);
     
    System.out.println();
     
    printPoints(q, 5, 0);
}
}
 
// This code is contributed by Rajnis09

C#

// C# program to find the points on 
// a line of slope M at distance L
using System;
 
class GFG{
 
// Class to represent a co-ordinate
// point
public class Point
{
    public float x, y;
     
    public Point()
    {
        x = y = 0;
    }
     
    public Point(float a, float b)
    {
        x = a;
        y = b;
    }
};
 
// Function to print pair of points at
// distance 'l' and having a slope 'm'
// from the source
static void printPoints(Point source,
                        float l, int m)
{
     
    // m is the slope of line, and the
    // required Point lies distance l
    // away from the source Point
    Point a = new Point();
    Point b = new Point();
     
    // Slope is 0
    if (m == 0)
    {
        a.x = source.x + l;
        a.y = source.y;
 
        b.x = source.x - l;
        b.y = source.y;
    }
 
    // If slope is infinite
    else if (Double.IsInfinity(m))
    {
        a.x = source.x;
        a.y = source.y + l;
 
        b.x = source.x;
        b.y = source.y - l;
    }
    else
    {
        float dx = (float)(l / Math.Sqrt(
                           1 + (m * m)));
        float dy = m * dx;
        a.x = source.x + dx;
        a.y = source.y + dy;
        b.x = source.x - dx;
        b.y = source.y - dy;
    }
 
    // Print the first Point
    Console.WriteLine(a.x + ", " + a.y);
 
    // Print the second Point
    Console.WriteLine(b.x + ", " + b.y);
}
 
// Driver code
public static void Main(String[] args)
{
    Point p = new Point(2, 1),
          q = new Point(1, 0);
           
    printPoints(p, (float)Math.Sqrt(2), 1);
     
    Console.WriteLine();
     
    printPoints(q, 5, 0);
}
}
 
// This code is contributed by Amit Katiyar

输出：

3, 2
1, 0

6, 0
-4, 0