给定两个坐标,找到一条直线的斜率。
例子:
Input : x1 = 4, y1 = 2,
x2 = 2, y2 = 5
Output : Slope is -1.5
方法:要计算一条线的斜率,您只需要从该线开始计算两个点,即(x1,y1)和(x2,y2)。用于从两个点计算斜率的公式为:
下面是上述方法的实现:
c++
// C program for slope of line
#include
using namespace std;
// function to find the slope of a straight line
float slope(float x1, float y1, float x2, float y2)
{
return (y2 - y1) / (x2 - x1);
}
// driver code to check the above function
int main()
{
float x1 = 4, y1 = 2;
float x2 = 2, y2 = 5;
cout << "Slope is: "
<< slope(x1, y1, x2, y2);
return 0;
}
Java
// Java program for slope of line
import java.io.*;
class GFG {
static float slope(float x1, float y1,
float x2, float y2)
{
return (y2 - y1) / (x2 - x1);
}
public static void main(String[] args)
{
float x1 = 4, y1 = 2;
float x2 = 2, y2 = 5;
System.out.println("Slope is " +
slope(x1, y1, x2, y2));
}
}
Python
# Python program for slope of line
def slope(x1, y1, x2, y2):
return (float)(y2-y1)/(x2-x1)
# driver code
x1 = 4
y1 = 2
x2 = 2
y2 = 5
print "Slope is :", slope(x1, y1, x2, y2)
C#
// C# program for slope of line
using System;
class GFG
{
static float slope(float x1, float y1,
float x2, float y2)
{
return (y2 - y1) / (x2 - x1);
}
// Driver code
public static void Main()
{
float x1 = 4, y1 = 2;
float x2 = 2, y2 = 5;
Console.WriteLine("Slope is " +
slope(x1, y1, x2, y2));
}
}
// This code is contributed by vt_m.
PHP
Javascript
输出:
Slope is: -1.5
时间复杂度: O(1)
辅助空间: O(1)