Check if three straight lines are concurrent or not

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 : No

Let
a1x + b1y + c1 = 0 ………. (1)
a2x + b2y + c2 = 0 ………. (2)
a3x + b3y + c3 = 0 ………. (3)

Suppose the eqn (i) and (ii) intersets at (x1, y1). Then (x1, y1) will satisfy bothe 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 coodinates 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 implemenatation of this approach:

C++

 // CPP Program to check if three straight  // line are concurrent or not #include using namespace std;    // Return true if three line are concurrent, // else false. 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); }    // Driven Program 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; }

Java

 // Java Program to check if three straight  // line are concurrent or no import java.io.*;    class GFG {        // Return true if three line are concurrent,     // else false.     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);     }            // Driven Program     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");     } }    // This code is contributed by anuj_67.

Python 3

 # Python3 Program to check if three straight  # line are concurrent or not    # Return true if three line are concurrent, # else false. 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)       # Driven Program 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")    # This code is contributed by Smitha

C#

 // C# Program to check if three straight  // line are concurrent or no using System;    class GFG {        // Return true if three line are concurrent,     // else false.     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);     }            // Driven Program     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");     } }    // This code is contributed by anuj_67.

PHP



Output:

Yes

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