给定直线(m1,m2)的中点和直线的一个坐标(x1,y1),找到直线的另一个端点(x2,y2)
例子:
Input : x1 = –1, y1 = 2, and
m1 = 3, m2 = –6
Output : x2 = 7, y2 = 10
Input : x1 = 6.4, y1 = 3 and
m1 = –10.7, m2 = 4
Output : x2 = 3, y2 = 4
中点公式:(x1,y2)和(x2,y2)这两个点的中点是使用以下公式找到的点M:
M =((x1 + x2)/ 2,(y1 + y2)/ 2),
我们需要一个(x2,y2),所以我们修改公式
m1 = ((x1+x2)/2), m2 = ((y1+y2)/2)
2*m1 = (x1+x2), 2*m2 = (y1+y2)
x2 = (2*m1 - x1), y2 = (2*m2 - y1)
C++
// CPP program to find the end point of a line
#include
using namespace std;
// CPP function to find the end point of a line
void otherEndPoint(int x1, int y1, int m1, int m2)
{
// find end point for x cordinates
float x2 = (float)(2 * m1 - x1);
// find end point for y cordinates
float y2 = (float)(2 * m2 - y1);
cout << "x2 = " << x2 << ", "
<< "y2 = " << y2;
}
// Driven Program
int main()
{
int x1 = -4, y1 = -1, m1 = 3, m2 = 5;
otherEndPoint(x1, y1, m1, m2);
return 0;
}
Java
// Java program to find the end point of a line
class GFG {
// CPP function to find the end point
// of a line
static void otherEndPoint(int x1, int y1,
int m1, int m2)
{
// find end point for x cordinates
float x2 = (float)(2 * m1 - x1);
// find end point for y cordinates
float y2 = (float)(2 * m2 - y1);
System.out.println("x2 = " + x2 + ", "
+ "y2 = " + y2);
}
// Driven Program
public static void main(String args[])
{
int x1 = -4, y1 = -1, m1 = 3, m2 = 5;
otherEndPoint(x1, y1, m1, m2);
}
}
// This code is contributed by JaideepPyne.
Python3
# Python3 program to find the end
# point of a line
# function to find the end point
# of a line
def otherEndPoint(x1, y1, m1, m2):
# find end point for x cordinates
x2 = (2 * m1 - x1)
# find end point for y cordinates
y2 = (2 * m2 - y1)
print ("x2 = {}, y2 = {}"
. format(x2, y2))
# Driven Program
x1 = -4
y1 = -1
m1 = 3
m2 = 5
otherEndPoint(x1, y1, m1, m2)
# This code is contributed by
# Manish Shaw (manishshaw1)
C#
// C# program to find the
// end point of a line
using System;
class GFG {
// function to find the
// end pointof a line
static void otherEndPoint(int x1, int y1,
int m1, int m2)
{
// find end point for x cordinates
float x2 = (float)(2 * m1 - x1);
// find end point for y cordinates
float y2 = (float)(2 * m2 - y1);
Console.WriteLine("x2 = " + x2 + ", "
+ "y2 = " + y2);
}
// Driver Program
public static void Main(String []args)
{
int x1 = -4, y1 = -1, m1 = 3, m2 = 5;
otherEndPoint(x1, y1, m1, m2);
}
}
// This code is contributed by nitin mittal.
PHP
输出:
x2 = 10, y2 = 11
时间复杂度: O(1)
辅助空间: O(1)