Area of triangle formed by the axes of co-ordinates and a given straight line
Last Updated :
07 Jun, 2022
Given a straight line with equation coefficients as a, b & c(ax + by + c = 0), the task is to find the area of the triangle formed by the axes of co-ordinates and this straight line.
Examples:
Input: a = -2, b = 4, c = 3
Output: 0.5625
Input: a = 4, b = 3, c = 12
Output: 6
Approach:
- Let PQ be the straight line having AB, the line segment between the axes.
The equation is,
ax + by + c = 0
- so, in intercept form it can be expressed as,
x/(-c/a) + y/(-c/b) = 1
- So, the x-intercept = -c/a
the y-intercept = -c/b
- So, it is very clear now the base of the triangle AOB will be -c/a
and the base of the triangle AOB will be -c/b
- So, area of the triangle
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
double area( double a, double b, double c)
{
double d = fabs ((c * c) / (2 * a * b));
return d;
}
int main()
{
double a = -2, b = 4, c = 3;
cout << area(a, b, c);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static double area( double a, double b, double c)
{
double d = Math.abs((c * c) / ( 2 * a * b));
return d;
}
public static void main (String[] args)
{
double a = - 2 , b = 4 , c = 3 ;
System.out.println(area(a, b, c));
}
}
|
Python3
def area(a, b, c):
d = abs ((c * c) / ( 2 * a * b))
return d
a = - 2
b = 4
c = 3
print (area(a, b, c))
|
C#
using System;
class GFG
{
static double area( double a, double b, double c)
{
double d = Math.Abs((c * c) / (2 * a * b));
return d;
}
static public void Main ()
{
double a = -2, b = 4, c = 3;
Console.WriteLine (area(a, b, c));
}
}
|
PHP
<?php
function area( $a , $b , $c )
{
$d = abs (( $c * $c ) / (2 * $a * $b ));
return $d ;
}
$a = -2;
$b = 4;
$c = 3;
echo area( $a , $b , $c );
?>
|
Javascript
<script>
function area(a , b , c)
{
var d = Math.abs((c * c) / (2 * a * b));
return d;
}
var a = -2, b = 4, c = 3;
document.write(area(a, b, c));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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