Find points at a given distance on a line of given slope

Given the co-ordinates of a 2-dimensional point p(x₀, y₀). Find the points at a distance L away from it, such that the line formed by joining these points has a slope of M.

Examples:

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

We need to find two points that are L distance from given point, on a line with slope M.
The idea has been introduced in below post.
Find Corners of Rectangle using mid points

Based on the input slope, the problem can be classified into 3 categories.

If slope is zero, we just need to adjust the x coordinate of the source point
If slope is infinite, the we need to adjust the y coordinate
For other values of slope, we can use the following equations to find the points

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}}

Now using the above formula we can find the required points.

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

Python

# 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 math.isfinite(m) is False:
        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)

// 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

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>

Output

3, 2
1, 0

6, 0
-4, 0

Time Complexity: O(1)
Auxiliary Space: O(1)

This article is contributed by Ashutosh Kumar :D

Find points at a given distance on a line of given slope

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