Consider a rectangle ABCD, we’re given the co-ordinates of the mid points of side AD and BC (p and q respectively) along with their length L (AD = BC = L). Now given the parameters, we need to print the co-ordinates of the 4 points A, B, C and D.
Examples:
Input : p = (1, 0)
q = (1, 2)
L = 2
Output : (0, 0), (0, 2), (2, 2), (2, 0)
Explanation:
The printed points form a rectangle which
satisfy the input constraints.
Input : p = (1, 1)
q = (-1, -1)
L = 2*sqrt(2)
Output : (0, 2), (-2, 0), (0, -2), (2, 0)
From the problem statement 3 cases can arise :
- The Rectangle is horizontal i.e., AD and BC are parallel to X-axis
- The Rectangle is vertical i.e., AD and BC are parallel to Y-axis
- The Rectangle is inclined at a certain angle with the axes
The first two cases are trivial and can easily be solved using basic geometry. For the third case we need to apply some mathematical concepts to find the points.
Consider the above diagram for clarity. We have the co-ordinates of p and q. Thus we can find the slope of AD and BC (As pq is perpendicular to AD). Once we have the slope of AD, we can find the equation of straight line passing through AD. Now we can apply distance formula to obtain the displacements along X and Y axes.
If slope of AD = m, then m = (p.x- q.x)/(q.y - p.y) and displacement along X axis, dx = L/(2*sqrt(1+m*m)) Similarly, dy = m*L/(2*sqrt(1+m*m))
Now we can simply find the co-ordinates of 4 corners by simply adding and subtracting the displacements obtained accordingly.
Below is the implementation .
C++
// C++ program to find corner points of// a rectangle using given length and middle// points.#include <bits/stdc++.h>using namespace std;// Structure to represent a co-ordinate pointstruct Point{ float x, y; Point() { x = y = 0; } Point(float a, float b) { x = a, y = b; }};// This function receives two points and length// of the side of rectangle and prints the 4// corner points of the rectanglevoid printCorners(Point p, Point q, float l){ Point a, b, c, d; // horizontal rectangle if (p.x == q.x) { a.x = p.x - (l/2.0); a.y = p.y; d.x = p.x + (l/2.0); d.y = p.y; b.x = q.x - (l/2.0); b.y = q.y; c.x = q.x + (l/2.0); c.y = q.y; } // vertical rectangle else if (p.y == q.y) { a.y = p.y - (l/2.0); a.x = p.x; d.y = p.y + (l/2.0); d.x = p.x; b.y = q.y - (l/2.0); b.x = q.x; c.y = q.y + (l/2.0); c.x = q.x; } // slanted rectangle else { // calculate slope of the side float m = (p.x-q.x)/float(q.y-p.y); // calculate displacements along axes float dx = (l /sqrt(1+(m*m))) *0.5 ; float dy = m*dx; a.x = p.x - dx; a.y = p.y - dy; d.x = p.x + dx; d.y = p.y + dy; b.x = q.x - dx; b.y = q.y - dy; c.x = q.x + dx; c.y = q.y + dy; } cout << a.x << ", " << a.y << " n" << b.x << ", " << b.y << "n"; << c.x << ", " << c.y << " n" << d.x << ", " << d.y << "nn";}// Driver codeint main(){ Point p1(1, 0), q1(1, 2); printCorners(p1, q1, 2); Point p(1, 1), q(-1, -1); printCorners(p, q, 2*sqrt(2)); return 0;} |
Java
// Java program to find corner points of // a rectangle using given length and middle // points. class GFG { // Structure 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; } }; // This function receives two points and length // of the side of rectangle and prints the 4 // corner points of the rectangle static void printCorners(Point p, Point q, float l) { Point a = new Point(), b = new Point(), c = new Point(), d = new Point(); // horizontal rectangle if (p.x == q.x) { a.x = (float) (p.x - (l / 2.0)); a.y = p.y; d.x = (float) (p.x + (l / 2.0)); d.y = p.y; b.x = (float) (q.x - (l / 2.0)); b.y = q.y; c.x = (float) (q.x + (l / 2.0)); c.y = q.y; } // vertical rectangle else if (p.y == q.y) { a.y = (float) (p.y - (l / 2.0)); a.x = p.x; d.y = (float) (p.y + (l / 2.0)); d.x = p.x; b.y = (float) (q.y - (l / 2.0)); b.x = q.x; c.y = (float) (q.y + (l / 2.0)); c.x = q.x; } // slanted rectangle else { // calculate slope of the side float m = (p.x - q.x) / (q.y - p.y); // calculate displacements along axes float dx = (float) ((l / Math.sqrt(1 + (m * m))) * 0.5); float dy = m * dx; a.x = p.x - dx; a.y = p.y - dy; d.x = p.x + dx; d.y = p.y + dy; b.x = q.x - dx; b.y = q.y - dy; c.x = q.x + dx; c.y = q.y + dy; } System.out.print((int)a.x + ", " + (int)a.y + " \n" + (int)b.x + ", " + (int)b.y + "\n" + (int)c.x + ", " + (int)c.y + " \n" + (int)d.x + ", " + (int)d.y + "\n"); } // Driver code public static void main(String[] args) { Point p1 = new Point(1, 0), q1 = new Point(1, 2); printCorners(p1, q1, 2); Point p = new Point(1, 1), q = new Point(-1, -1); printCorners(p, q, (float) (2 * Math.sqrt(2))); }}// This code contributed by Rajput-Ji |
Python3
# Python3 program to find corner points of# a rectangle using given length and middle# points.import math# Structure to represent a co-ordinate pointclass Point: def __init__(self, a = 0, b = 0): self.x = a self.y = b # This function receives two points and length# of the side of rectangle and prints the 4# corner points of the rectangledef printCorners(p, q, l): a, b, c, d = Point(), Point(), Point(), Point() # Horizontal rectangle if (p.x == q.x): a.x = p.x - (l / 2.0) a.y = p.y d.x = p.x + (l / 2.0) d.y = p.y b.x = q.x - (l / 2.0) b.y = q.y c.x = q.x + (l / 2.0) c.y = q.y # Vertical rectangle else if (p.y == q.y): a.y = p.y - (l / 2.0) a.x = p.x d.y = p.y + (l / 2.0) d.x = p.x b.y = q.y - (l / 2.0) b.x = q.x c.y = q.y + (l / 2.0) c.x = q.x # Slanted rectangle else: # Calculate slope of the side m = (p.x - q.x) / (q.y - p.y) # Calculate displacements along axes dx = (l / math.sqrt(1 + (m * m))) * 0.5 dy = m * dx a.x = p.x - dx a.y = p.y - dy d.x = p.x + dx d.y = p.y + dy b.x = q.x - dx b.y = q.y - dy c.x = q.x + dx c.y = q.y + dy print(int(a.x), ", ", int(a.y), sep = "") print(int(b.x), ", ", int(b.y), sep = "") print(int(c.x), ", ", int(c.y), sep = "") print(int(d.x), ", ", int(d.y), sep = "") print() # Driver codep1 = Point(1, 0)q1 = Point(1, 2)printCorners(p1, q1, 2)p = Point(1, 1)q = Point(-1, -1)printCorners(p, q, 2 * math.sqrt(2))# This code is contributed by shubhamsingh10 |
C#
// C# program to find corner points of // a rectangle using given length and middle // points. using System; class GFG { // Structure 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; } }; // This function receives two points and length // of the side of rectangle and prints the 4 // corner points of the rectangle static void printCorners(Point p, Point q, float l) { Point a = new Point(), b = new Point(), c = new Point(), d = new Point(); // horizontal rectangle if (p.x == q.x) { a.x = (float) (p.x - (l / 2.0)); a.y = p.y; d.x = (float) (p.x + (l / 2.0)); d.y = p.y; b.x = (float) (q.x - (l / 2.0)); b.y = q.y; c.x = (float) (q.x + (l / 2.0)); c.y = q.y; } // vertical rectangle else if (p.y == q.y) { a.y = (float) (p.y - (l / 2.0)); a.x = p.x; d.y = (float) (p.y + (l / 2.0)); d.x = p.x; b.y = (float) (q.y - (l / 2.0)); b.x = q.x; c.y = (float) (q.y + (l / 2.0)); c.x = q.x; } // slanted rectangle else { // calculate slope of the side float m = (p.x - q.x) / (q.y - p.y); // calculate displacements along axes float dx = (float) ((l / Math.Sqrt(1 + (m * m))) * 0.5); float dy = m * dx; a.x = p.x - dx; a.y = p.y - dy; d.x = p.x + dx; d.y = p.y + dy; b.x = q.x - dx; b.y = q.y - dy; c.x = q.x + dx; c.y = q.y + dy; } Console.Write((int)a.x + ", " + (int)a.y + " \n" + (int)b.x + ", " + (int)b.y + "\n" + (int)c.x + ", " + (int)c.y + " \n" + (int)d.x + ", " + (int)d.y + "\n"); } // Driver code public static void Main(String[] args) { Point p1 = new Point(1, 0), q1 = new Point(1, 2); printCorners(p1, q1, 2); Point p = new Point(1, 1), q = new Point(-1, -1); printCorners(p, q, (float) (2 * Math.Sqrt(2))); } } // This code has been contributed by 29AjayKumar |
Javascript
<script>// Javascript program to find corner points of // a rectangle using given length and middle // points. // Structure to represent a co-ordinate point class Point { constructor(a,b) { this.x=a; this.y=b; }}// This function receives two points and length // of the side of rectangle and prints the 4 // corner points of the rectangle function printCorners(p,q,l){ let a = new Point(), b = new Point(), c = new Point(), d = new Point(); // horizontal rectangle if (p.x == q.x) { a.x = (p.x - (l / 2.0)); a.y = p.y; d.x = (p.x + (l / 2.0)); d.y = p.y; b.x = (q.x - (l / 2.0)); b.y = q.y; c.x = (q.x + (l / 2.0)); c.y = q.y; } // vertical rectangle else if (p.y == q.y) { a.y = (p.y - (l / 2.0)); a.x = p.x; d.y = (p.y + (l / 2.0)); d.x = p.x; b.y = (q.y - (l / 2.0)); b.x = q.x; c.y = (q.y + (l / 2.0)); c.x = q.x; } // slanted rectangle else { // calculate slope of the side let m = (p.x - q.x) / (q.y - p.y); // calculate displacements along axes let dx = ((l / Math.sqrt(1 + (m * m))) * 0.5); let dy = m * dx; a.x = p.x - dx; a.y = p.y - dy; d.x = p.x + dx; d.y = p.y + dy; b.x = q.x - dx; b.y = q.y - dy; c.x = q.x + dx; c.y = q.y + dy; } document.write(a.x + ", " + a.y + " <br>" + b.x + ", " + b.y + "<br>" + c.x + ", " + c.y + " <br>" + d.x + ", " + d.y + "<br>");}// Driver code let p1 = new Point(1, 0), q1 = new Point(1, 2);printCorners(p1, q1, 2);let p = new Point(1, 1), q = new Point(-1, -1);printCorners(p, q, (2 * Math.sqrt(2)));// This code is contributed by rag2127</script> |
Output:
0, 0 0, 2 2, 2 2, 0 0, 2 -2, 0 0, -2 2, 0
Time Complexity: O(1)
Auxiliary Space: O(1)
Reference:
StackOverflow
This article is contributed by Aarti_Rathi and Ashutosh Kumar 😀 If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
