给定三点,检查它们是否位于直线(共线)上
例子 :
Input : (1, 1), (1, 4), (1, 5)
Output : Yes
The points lie on a straight line
Input : (1, 5), (2, 5), (4, 6)
Output : No
The points do not lie on a straight line
第一种方法
如果由这三个点的三角形形成的面积为零,则三个点位于直线上。因此,我们将检查由三角形形成的面积是否为零
Formula for area of triangle is :
0.5 * [x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)]
The formula is basically half of determinant
value of following.
x1 x2 x3
y1 y2 y3
1 1 1
The above formula is derived from shoelace formula.
C++
// C++ program to check if three
// points are collinear or not
// using area of triangle.
#include
#include
#include
using namespace std;
// function to check if point
// collinear or not
void collinear(int x1, int y1, int x2,
int y2, int x3, int y3)
{
// Calculation the area of
// triangle. We have skipped
// multiplication with 0.5
// to avoid floating point
// computations
int a = x1 * (y2 - y3) +
x2 * (y3 - y1) +
x3 * (y1 - y2);
if (a == 0)
cout << "Yes";
else
cout << "No";
}
// Driver Code
int main()
{
int x1 = 1, x2 = 1, x3 = 1,
y1 = 1, y2 = 4, y3 = 5;
collinear(x1, y1, x2, y2, x3, y3);
return 0;
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
C
// C program to check if three
// points are collinear or not
// using area of triangle.
#include
#include
#include
// function to check if point
// collinear or not
void collinear(int x1, int y1, int x2,
int y2, int x3, int y3)
{
// Calculation the area of
// triangle. We have skipped
// multiplication with 0.5
// to avoid floating point
// computations
int a = x1 * (y2 - y3) +
x2 * (y3 - y1) +
x3 * (y1 - y2);
if (a == 0)
printf("Yes");
else
printf("No");
}
// Driver Code
int main()
{
int x1 = 1, x2 = 1, x3 = 1,
y1 = 1, y2 = 4, y3 = 5;
collinear(x1, y1, x2, y2, x3, y3);
return 0;
}
Java
// Java program to check if
// three points are collinear
// or not using area of triangle.
class GFG
{
// function to check if
// point collinear or not
static void collinear(int x1, int y1, int x2,
int y2, int x3, int y3)
{
/* Calculation the area of
triangle. We have skipped
multiplication with 0.5
to avoid floating point
computations */
int a = x1 * (y2 - y3) +
x2 * (y3 - y1) +
x3 * (y1 - y2);
if (a == 0)
System.out.println("Yes");
else
System.out.println("No");
}
// Driver Code
public static void main(String args[])
{
int x1 = 1, x2 = 1, x3 = 1,
y1 = 1, y2 = 4, y3 = 5;
collinear(x1, y1, x2, y2, x3, y3);
}
}
// This code is contributed by Sam007.
Python
# Python program to check
# if three points are collinear
# or not using area of triangle.
# function to check if
# point collinear or not
def collinear(x1, y1, x2, y2, x3, y3):
""" Calculation the area of
triangle. We have skipped
multiplication with 0.5 to
avoid floating point computations """
a = x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)
if (a == 0):
print "Yes"
else:
print "No"
# Driver Code
x1, x2, x3, y1, y2, y3 = 1, 1, 1, 1, 4, 5
collinear(x1, y1, x2, y2, x3, y3)
# This code is contributed
# by Sachin Bisht
C#
// C# program to check if
// three points are collinear
// or not using area of triangle.
using System;
class GFG
{
/* function to check if
point collinear or not */
static void collinear(int x1, int y1, int x2,
int y2, int x3, int y3)
{
/* Calculation the area of
triangle. We have skipped
multiplication with 0.5 to
avoid floating point computations */
int a = x1 * (y2 - y3) +
x2 * (y3 - y1) +
x3 * (y1 - y2);
if (a == 0)
Console.Write("Yes");
else
Console.Write("No");
}
// Driver code
public static void Main ()
{
int x1 = 1, x2 = 1, x3 = 1,
y1 = 1, y2 = 4, y3 = 5;
collinear(x1, y1, x2, y2, x3, y3);
}
}
// This code is contributed by Sam007.
PHP
C
// Slope based solution to check
// if three points are collinear.
#include
#include
/* function to check if
point collinear or not*/
void collinear(int x1, int y1, int x2,
int y2, int x3, int y3)
{
if ((y3 - y2) * (x2 - x1) ==
(y2 - y1) * (x3 - x2))
printf("Yes");
else
printf("No");
}
// Driver Code
int main()
{
int x1 = 1, x2 = 1, x3 = 0,
y1 = 1, y2 = 6, y3 = 9;
collinear(x1, y1, x2, y2, x3, y3);
return 0;
}
Java
// Slope based solution to check
// if three points are collinear.
import java.io.*;
class GFG {
/* function to check if
point collinear or not*/
static void cool_line(int x1, int y1, int x2,
int y2, int x3, int y3)
{
if ((y3 - y2) * (x2 - x1) ==
(y2 - y1) * (x3 - x2))
System.out.println("Yes");
else
System.out.println("No");
}
// Driver Code
public static void main (String[] args) {
int a1 = 1, a2 = 1, a3 = 0,
b1 = 1, b2 = 6, b3 = 9;
cool_line(a1, b1, a2, b2, a3, b3);
}
}
//This Code is Contributed by ajit
Python
# Slope based solution to check if three
# points are collinear.
# function to check if
# point collinear or not
def collinear(x1, y1, x2, y2, x3, y3):
if ((y3 - y2)*(x2 - x1) == (y2 - y1)*(x3 - x2)):
print ("Yes")
else:
print ("No")
# Driver Code
x1, x2, x3, y1, y2, y3 = 1, 1, 0, 1, 6, 9
collinear(x1, y1, x2, y2, x3, y3);
# This code is contributed
# by Sachin Bisht
C#
// Slope based solution to check
// if three points are collinear.
using System;
class GFG
{
/* function to check if
point collinear or not*/
static void cool_line(int x1, int y1, int x2,
int y2, int x3, int y3)
{
if ((y3 - y2) * (x2 - x1) ==
(y2 - y1) * (x3 - x2))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
// Driver Code
static public void Main ()
{
int a1 = 1, a2 = 1, a3 = 0,
b1 = 1, b2 = 6, b3 = 9;
cool_line(a1, b1, a2, b2, a3, b3);
}
}
// This code is contributed by ajit
PHP
Javascript
输出 :
Yes
第二种方法
For three points, slope of any pair of points
must be same as other pair.
For example, slope of line joining (x2, y2)
and (x3, y3), and line joining (x1, y1) and
(x2, y2) must be same.
(y3 - y2)/(x3 - x2) = (y2 - y1)/(x2 - x1)
In other words,
(y3 - y2)(x2 - x1) = (y2 - y1)(x3 - x2)
如果等于零,则点位于一条直线上
C
// Slope based solution to check
// if three points are collinear.
#include
#include
/* function to check if
point collinear or not*/
void collinear(int x1, int y1, int x2,
int y2, int x3, int y3)
{
if ((y3 - y2) * (x2 - x1) ==
(y2 - y1) * (x3 - x2))
printf("Yes");
else
printf("No");
}
// Driver Code
int main()
{
int x1 = 1, x2 = 1, x3 = 0,
y1 = 1, y2 = 6, y3 = 9;
collinear(x1, y1, x2, y2, x3, y3);
return 0;
}
Java
// Slope based solution to check
// if three points are collinear.
import java.io.*;
class GFG {
/* function to check if
point collinear or not*/
static void cool_line(int x1, int y1, int x2,
int y2, int x3, int y3)
{
if ((y3 - y2) * (x2 - x1) ==
(y2 - y1) * (x3 - x2))
System.out.println("Yes");
else
System.out.println("No");
}
// Driver Code
public static void main (String[] args) {
int a1 = 1, a2 = 1, a3 = 0,
b1 = 1, b2 = 6, b3 = 9;
cool_line(a1, b1, a2, b2, a3, b3);
}
}
//This Code is Contributed by ajit
Python
# Slope based solution to check if three
# points are collinear.
# function to check if
# point collinear or not
def collinear(x1, y1, x2, y2, x3, y3):
if ((y3 - y2)*(x2 - x1) == (y2 - y1)*(x3 - x2)):
print ("Yes")
else:
print ("No")
# Driver Code
x1, x2, x3, y1, y2, y3 = 1, 1, 0, 1, 6, 9
collinear(x1, y1, x2, y2, x3, y3);
# This code is contributed
# by Sachin Bisht
C#
// Slope based solution to check
// if three points are collinear.
using System;
class GFG
{
/* function to check if
point collinear or not*/
static void cool_line(int x1, int y1, int x2,
int y2, int x3, int y3)
{
if ((y3 - y2) * (x2 - x1) ==
(y2 - y1) * (x3 - x2))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
// Driver Code
static public void Main ()
{
int a1 = 1, a2 = 1, a3 = 0,
b1 = 1, b2 = 6, b3 = 9;
cool_line(a1, b1, a2, b2, a3, b3);
}
}
// This code is contributed by ajit
的PHP
Java脚本
输出 :
No