Find the coordinates of a triangle whose Area = (S / 2)
Given an integer S, the task is to find the coordinates of a triangle whose area is (S / 2).
Examples:
Input: S = 4
Output:
(0, 0)
(1000000000, 1)
(999999996, 1)
Input: S = 15
Output:
(0, 0)
(1000000000, 1)
(999999985, 1)
Approach:
- It is known than the area of the triangle whose coordinates are (X1, Y1), (X2, Y2) and (X3, Y3) is given by A = ((X1 * Y2) + (X2 * Y3) + (X3 * Y1) – (X1 * Y3) – (X2 * Y1) – (X3 * Y2)) / 2.
- Now fixing (X1, Y1) to (0, 0) gives A = ((X2 * Y3) – (X3 * Y2)) / 2.
- It is given that A = S / 2 which implies S = (X2 * Y3) – (X3 * Y2).
- Now fix (X2, Y2) to (109, 1) and the equation becomes S = 109 * Y3 – X3 which can be solved by taking an integer value of a variable that given the integer value for the other variable.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
const long MAX = 1000000000;
void findTriangle( long S)
{
long X1 = 0, Y1 = 0;
long X2 = MAX, Y2 = 1;
long X3 = (MAX - S % MAX) % MAX;
long Y3 = (S + X3) / MAX;
cout << "(" << X1 << ", " << Y1 << ")\n" ;
cout << "(" << X2 << ", " << Y2 << ")\n" ;
cout << "(" << X3 << ", " << Y3 << ")" ;
}
int main()
{
long S = 4;
findTriangle(S);
return 0;
}
|
Java
class GFG
{
static final long MAX = 1000000000 ;
static void findTriangle( long S)
{
long X1 = 0 , Y1 = 0 ;
long X2 = MAX, Y2 = 1 ;
long X3 = (MAX - S % MAX) % MAX;
long Y3 = (S + X3) / MAX;
System.out.println( "(" + X1 +
", " + Y1 + ")" );
System.out.println( "(" + X2 +
", " + Y2 + ")" );
System.out.println( "(" + X3 +
", " + Y3 + ")" );
}
public static void main (String[] args)
{
long S = 4 ;
findTriangle(S);
}
}
|
Python3
MAX = 1000000000 ;
def findTriangle(S) :
X1 = 0 ; Y1 = 0 ;
X2 = MAX ; Y2 = 1 ;
X3 = ( MAX - S % MAX ) % MAX ;
Y3 = (S + X3) / MAX ;
print ( "(" , X1, "," , Y1, ")" );
print ( "(" , X2, "," , Y2, ")" );
print ( "(" , X3, "," , Y3, ")" );
if __name__ = = "__main__" :
S = 4 ;
findTriangle(S);
|
C#
using System;
class GFG
{
static readonly long MAX = 1000000000;
static void findTriangle( long S)
{
long X1 = 0, Y1 = 0;
long X2 = MAX, Y2 = 1;
long X3 = (MAX - S % MAX) % MAX;
long Y3 = (S + X3) / MAX;
Console.WriteLine( "(" + X1 +
", " + Y1 + ")" );
Console.WriteLine( "(" + X2 +
", " + Y2 + ")" );
Console.WriteLine( "(" + X3 +
", " + Y3 + ")" );
}
public static void Main (String[] args)
{
long S = 4;
findTriangle(S);
}
}
|
Javascript
<script>
let MAX = 1000000000;
function findTriangle( S)
{
let X1 = 0, Y1 = 0;
let X2 = MAX, Y2 = 1;
let X3 = (MAX - S % MAX) % MAX;
let Y3 = (S + X3) / MAX;
document.write( "(" + X1 + ", " + Y1 + ")<br/>" );
document.write( "(" + X2 + ", " + Y2 + ")<br/>" );
document.write( "(" + X3 + ", " + Y3 + ")<br/>" )
}
let S = 4;
findTriangle(S);
</script>
|
Output:
(0, 0)
(1000000000, 1)
(999999996, 1)
Time Complexity: O(1)
Auxiliary Space: O(1)
Last Updated :
24 Feb, 2023
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