Given the co-ordinates of a 2-dimensional point p(x_{0}, y_{0}). 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

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 infinte` ` ` `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 infinte` ` ` `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 infinte` ` ` `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` |

**Output: **

3, 2 1, 0 6, 0 -4, 0

This article is contributed by **Ashutosh Kumar** π If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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