Find points at a given distance on a line of given slope (original) (raw)

Last Updated : 23 Jul, 2025

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#

// 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

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