Given three lines equation,
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
a3x + b3y + c3 = 0
The task is to check whether the given three lines are concurrent or not. Three straight lines are said to be concurrent if they pass through a point i.e., they meet at a point.
Examples:
Input : a1 = 2, b1 = -3, c1 = 5
a2 = 3, b2 = 4, c2 = -7
a3 = 9, b3 = -5, c3 = 8
Output : Yes
Input : a1 = 2, b1 = -3, c1 = 5
a2 = 3, b2 = 4, c2 = -7
a3 = 9, b3 = -5, c3 = 4
Output : NoLet
a1x + b1y + c1 = 0 ………. (1)
a2x + b2y + c2 = 0 ………. (2)
a3x + b3y + c3 = 0 ………. (3)
Suppose the eqn (i) and (ii) intersects at (x1, y1). Then (x1, y1) will satisfy both the equations.
Therefore, solving (i) and (ii) using method of cross-multiplication, we get,
(x1/b1c2 – b2c1) = (y1/c1a2 – c2a1) = (1/a1b2 – a2b1)
Therefore,
x1 = (b1c2 – b2c1/a1b2 – a2b1) and
y1 = (c1a2 – c2a1/a1b2 – a2b1), a1b2 – a2b1 != 0
Therefore, the required coordinates of the point of intersection of the lines (i) and (ii) are
(b1c2 – b2c1/a1b2 – a2b1, c1a2 – c2a1/a1b2 – a2b1)
For, three of line to be concurrent, (x1, y1) must satisfy the equation (iii) as well.
So,
a3x + b3y + c3 = 0
=> a3(b1c2 – b2c1/a1b2 – a2b1) + b3(c1a2 – c2a1/a1b2 – a2b1) + c3 = 0
=> a3(b1c2 – b2c1) + b3(c1a2 – c2a1) + c3(a1b2 – a2b1) = 0
So, we only need to check if above condition satisfy or not.
Below is the implementation of this approach:
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C++
#include <bits/stdc++.h>
using namespace std;
bool checkConcurrent(int a1, int b1, int c1,
int a2, int b2, int c2,
int a3, int b3, int c3)
{
return (a3 * (b1 * c2 - b2 * c1) +
b3 * (c1 * a2 - c2 * a1) +
c3 * (a1 * b2 - a2 * b1) == 0);
}
int main()
{
int a1 = 2, b1 = -3, c1 = 5;
int a2 = 3, b2 = 4, c2 = -7;
int a3 = 9, b3 = -5, c3 = 8;
(checkConcurrent(a1, b1, c1, a2, b2, c2,
a3, b3, c3) ? (cout << "Yes") : (cout << "No"));
return 0;
}
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Java
import java.io.*;
class GFG {
static boolean checkConcurrent(int a1, int b1,
int c1, int a2, int b2, int c2,
int a3, int b3, int c3)
{
return (a3 * (b1 * c2 - b2 * c1) +
b3 * (c1 * a2 - c2 * a1) +
c3 * (a1 * b2 - a2 * b1) == 0);
}
public static void main (String[] args)
{
int a1 = 2, b1 = -3, c1 = 5;
int a2 = 3, b2 = 4, c2 = -7;
int a3 = 9, b3 = -5, c3 = 8;
if(checkConcurrent(a1, b1, c1, a2, b2,
c2, a3, b3, c3))
System.out.println( "Yes");
else
System.out.println( "No");
}
}
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Python 3
def checkConcurrent(a1, b1, c1, a2, b2, c2,
a3, b3, c3):
return (a3 * (b1 * c2 - b2 * c1) +
b3 * (c1 * a2 - c2 * a1) +
c3 * (a1 * b2 - a2 * b1) == 0)
a1 = 2
b1 = -3
c1 = 5
a2 = 3
b2 = 4
c2 = -7
a3 = 9
b3 = -5
c3 = 8
if(checkConcurrent(a1, b1, c1, a2, b2, c2,
a3, b3, c3)):
print("Yes")
else:
print("No")
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C#
using System;
class GFG {
static bool checkConcurrent(int a1, int b1,
int c1, int a2, int b2, int c2,
int a3, int b3, int c3)
{
return (a3 * (b1 * c2 - b2 * c1) +
b3 * (c1 * a2 - c2 * a1) +
c3 * (a1 * b2 - a2 * b1) == 0);
}
public static void Main ()
{
int a1 = 2, b1 = -3, c1 = 5;
int a2 = 3, b2 = 4, c2 = -7;
int a3 = 9, b3 = -5, c3 = 8;
if(checkConcurrent(a1, b1, c1, a2, b2,
c2, a3, b3, c3))
Console.WriteLine( "Yes");
else
Console.WriteLine( "No");
}
}
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PHP
<?php
function checkConcurrent($a1, $b1, $c1,
$a2, $b2, $c2,
$a3, $b3, $c3)
{
return ($a3 * ($b1 * $c2 - $b2 * $c1) +
$b3 * ($c1 * $a2 - $c2 * $a1) +
$c3 * ($a1 * $b2 - $a2 * $b1) == 0);
}
$a1 = 2; $b1 = -3; $c1 = 5;
$a2 = 3; $b2 = 4; $c2 = -7;
$a3 = 9; $b3 = -5; $c3 = 8;
if(checkConcurrent($a1, $b1, $c1, $a2, $b2,
$c2, $a3, $b3, $c3))
echo "Yes";
else
echo "No";
?>
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Javascript
<script>
function checkConcurrent( a1, b1, c1,
a2, b2, c2,
a3, b3, c3)
{
return (a3 * (b1 * c2 - b2 * c1) +
b3 * (c1 * a2 - c2 * a1) +
c3 * (a1 * b2 - a2 * b1) == 0);
}
a1 = 2, b1 = -3, c1 = 5;
a2 = 3, b2 = 4, c2 = -7;
a3 = 9, b3 = -5, c3 = 8;
(checkConcurrent(a1, b1, c1, a2, b2, c2,
a3, b3, c3) ? (document.write( "Yes")) : (document.write( "No")));
</script>
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