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, 0We 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++
#include <bits/stdc++.h>
using namespace std;
struct Point {
float x, y;
Point() { x = y = 0; }
Point(float a, float b) { x = a, y = b; }
};
void printPoints(Point source, float l, int m)
{
Point a, b;
if (m == 0) {
a.x = source.x + l;
a.y = source.y;
b.x = source.x - l;
b.y = source.y;
}
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;
}
cout << a.x << ", " << a.y << endl;
cout << b.x << ", " << b.y << endl;
}
int main()
{
Point p(2, 1), q(1, 0);
printPoints(p, sqrt(2), 1);
cout << endl;
printPoints(q, 5, 0);
return 0;
}
|
Java
class GFG {
static class Point {
float x, y;
Point() { x = y = 0; }
Point(float a, float b)
{
x = a;
y = b;
}
};
static void printPoints(Point source, float l, int m)
{
Point a = new Point();
Point b = new Point();
if (m == 0) {
a.x = source.x + l;
a.y = source.y;
b.x = source.x - l;
b.y = source.y;
}
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;
}
System.out.println(a.x + ", " + a.y);
System.out.println(b.x + ", " + b.y);
}
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);
}
}
|
C#
using System;
class GFG {
public class Point {
public float x, y;
public Point() { x = y = 0; }
public Point(float a, float b)
{
x = a;
y = b;
}
};
static void printPoints(Point source, float l, int m)
{
Point a = new Point();
Point b = new Point();
if (m == 0) {
a.x = source.x + l;
a.y = source.y;
b.x = source.x - l;
b.y = source.y;
}
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;
}
Console.WriteLine(a.x + ", " + a.y);
Console.WriteLine(b.x + ", " + b.y);
}
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);
}
}
|
Python3
import math
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def printPoints(source, l, m):
a = Point(0, 0)
b = Point(0, 0)
if m == 0:
a.x = source.x + l
a.y = source.y
b.x = source.x - l
b.y = source.y
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(f"{a.x}, {a.y}")
print(f"{b.x}, {b.y}")
p = Point(2, 1)
q = Point(1, 0)
printPoints(p, math.sqrt(2), 1)
print("\n")
printPoints(q, 5, 0)
|
Javascript
<script>
class Point
{
constructor(x, y)
{
this.x = x;
this.y = y;
}
}
function printPoints(source, l, m)
{
let a = new Point();
let b = new Point();
if (m == 0)
{
a.x = source.x + l;
a.y = source.y;
b.x = source.x - l;
b.y = source.y;
}
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;
}
document.write(a.x + ", " + a.y+"\n");
document.write(b.x + ", " + b.y+"\n");
}
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);
</script>
|
Output3, 2
1, 0
6, 0
-4, 0
Time Complexity: O(1)
Auxiliary Space: O(1)
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