给定三行方程，
a 1 x + b 1 y + c 1 = 0
a 2 x + b 2 y + c 2 = 0
a 3 x + b 3 y + c 3 = 0
任务是检查给定的三行是否并发。如果三个直线通过一个点，即它们在一个点处相遇，则被认为是并发的。
例子：
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
让
a 1 x + b 1 y + c 1 = 0………。 (1)
a 2 x + b 2 y + c 2 = 0…………。 (2)
a 3 x + b 3 y + c 3 = 0…………。 (3)
假设等式(i)和(ii)在(x 1 ，y 1 )处交集。然后(x 1 ，y 1 )将满足这两个方程。
因此，使用交叉乘法求解(i)和(ii)，我们得到，
(x 1 / b 1 c 2 – b 2 c 1 )=(y 1 / c 1 a 2 – c 2 a 1 )=(1 / a 1 b 2 – a 2 b 1 )
所以，
x 1 =(b 1 c 2 – b 2 c 1 / a 1 b 2 – a 2 b 1 )并且
y 1 =(c 1 a 2 – c 2 a 1 / a 1 b 2 – a 2 b 1 )，a 1 b 2 – a 2 b 1 ！= 0
因此，线(i)和(ii)的相交点的所需余数为
(b 1 c 2 – b 2 c 1 / a 1 b 2 – a 2 b 1 ，c 1 a 2 – c 2 a 1 / a 1 b 2 – a 2 b 1 )
对于要同时存在的三行，(x 1 ，y 1 )也必须满足方程式(iii)。
所以，
a 3 x + b 3 y + c 3 = 0
=> a 3 (b 1 c 2 – b 2 c 1 / a 1 b 2 – a 2 b 1 )+ b 3 (c 1 a 2 – c 2 a 1 / a 1 b 2 – a 2 b 1 )+ c 3 = 0
=> a 3 (b 1 c 2 – b 2 c 1 )+ b 3 (c 1 a 2 – c 2 a 1 )+ c 3 (a 1 b 2 – a 2 b 1 )= 0
因此，我们只需要检查以上条件是否满足。
以下是此方法的实现：
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 SmithaC#
// 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
Yes