Encuentra puntos a una distancia dada en una línea de pendiente dada

Posted on by Rudeus Greyrat

Dadas las coordenadas de un punto bidimensional p(x 0 , y 0 ). Encuentre los puntos a una distancia L de él, tal que la línea formada al unir estos puntos tenga una pendiente de M.

Ejemplos: 

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

Necesitamos encontrar dos puntos que estén a una distancia L del punto dado, en una línea con pendiente M.
La idea se ha presentado en la publicación a continuación. 
Encuentra esquinas de rectángulo usando puntos medios

Según la pendiente de entrada, el problema se puede clasificar en 3 categorías.  

  1. Si la pendiente es cero, solo necesitamos ajustar la coordenada x del punto de origen
  2. Si la pendiente es infinita, necesitamos ajustar la coordenada y
  3. Para otros valores de pendiente, podemos usar las siguientes ecuaciones para encontrar los puntos

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}}
 

Ahora usando la fórmula anterior podemos encontrar los puntos requeridos.

C++

// C++ program to find the points on a line of
// slope M at distance L
#include <bits/stdc++.h>
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<float>
                                 ::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

Python3

# Python program to find the points on a line of
# slope M at distance L
import math
 
 
# structure to represent a co-ordinate
# point
 
 
class Point:
    def __init__(self, x, y):
        self.x = x
        self.y = y
 
# Function to print pair of points at
# distance 'l' and having a slope 'm'
# from the source
 
 
def printPoints(source, l, m):
    # m is the slope of line, and the
    # required Point lies distance l
    # away from the source Point
    a = Point(0, 0)
    b = Point(0, 0)
 
    # 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
    elif m != math.isfinite(m):
        a.x = source.x
        a.y = source.y + l
 
        b.x = source.x
        b.y = source.y - l
    else:
        dx = (l / math.sqrt(1 + (m * m)))
        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
    print(f"{a.x}, {a.y}")
 
    # print the second Point
    print(f"{b.x}, {b.y}")
 
 
# driver function
p = Point(2, 1)
q = Point(1, 0)
printPoints(p, math.sqrt(2), 1)
print("\n")
printPoints(q, 5, 0)
 
# The code is contributed by Gautam goel(gautamgoel962)

Javascript

<script>
    // Javascript program to find the points on
    // a line of slope M at distance L
       
    // Class to represent a co-ordinate
    // point
    class Point
    {
        constructor(x, y)
        {
            this.x = x;
            this.y = y;
        }
    }
     
    // Function to print pair of points at
    // distance 'l' and having a slope 'm'
    // from the source
    function printPoints(source, l, m)
    {
     
        // m is the slope of line, and the
        // required Point lies distance l
        // away from the source Point
        let a = new Point();
        let 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 (!isFinite(m))
        {
            a.x = source.x;
            a.y = source.y + l;
      
            b.x = source.x;
            b.y = source.y - l;
        }
        else
        {
            var dx = (l / Math.sqrt(1 + (m * m)));
            var 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
        document.write(a.x + ", " + a.y+"\n");
      
        // Print the second Point
        document.write(b.x + ", " + b.y+"\n");
    }
     
    // Driver code
    let p = new Point(2, 1);
    let q = new Point(1, 0);
    printPoints(p, Math.sqrt(2), 1);
    document.write("\n");
    printPoints(q, 5, 0);
     
    // This code is contributed by shruti456rawal
</script>

Producción: 

3, 2
1, 0

6, 0
-4, 0

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)

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Publicación traducida automáticamente

Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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