bladekiller97 The midPoint formula is the (first coordinate + second coordinate) divided by 2. So for example to find the mid-point between R(2, 2) and P(0, 0) you have to use the formula separately for X and Y. So the middle for X is (Mx = (Rx + Px) / 2 ) and for Y is (My = (Ry + Py) / 2)
Mx = (2 + 0) / 2 = 2 / 2 = 1
My = (2 + 0) / 2 = 2 / 2 = 1
so MidPoint is at coordinates M(1, 1).
What if the MidPoint is given and we have to calculate the point at the other half of the line. Then we use the formula above. So for example if we are given P(0, 0) and the midpoint M(1, 1) you should find R(x, y).
Mx = (Rx + Px) / 2 => Rx = 2Mx - Px
My = (Ry + Py) / 2 => Ry = 2My - Py
And thats how you find the coordinates for R. Here is my code :
#include <iostream> using namespace std; int main() { int n, px, py, mx, my; cin >> n; for (int i = 1; i <= n; i++){ cin >> px >> py >> mx >> my; int rx = 2 * mx - px; int ry = 2 * my - py; cout << rx << " " << ry << endl; } return 0; }
CrashWire Thank you for this information, highly helpful and I might've not done the challenge without this. Been many years without math. Took me a bit to get used to thinking again ^_^. You have my upvote.
utkarsh19972000 thanks for the help and logic i didnt able to understand the chellenge without your explenation
therawtechnoguy brilliant post, thank you for all the help
calcray96 Very simple in python.
for i in range(int(raw_input())): px, py, qx, qy = map(int, raw_input().split()) print ("{0} {1}".format(2*qx-px, 2*qy-py))
Altair007 hey, can u send me the complete solution . I am not able to import data. Thanks in advance.
linda_muemken That is the complete code. But i assume you try to use python3, there you have to change 'raw_input' into 'input', as this was renameD with python3.
pythonban Altair007, You dont need complete solution, coding is all about finding the solutions yourself, and a hint for solution: try section formula of co-ordinate geometry.
mkv_1724 python is beautiful language!
josef_klotzner Let me clarify please, that this solution might be simple for you, but it is definitly not a "simple" solution for a pyhton beginner. But it is a short solution and a clever solution, which needs quite some knowledge in python. I like it.
mrcoelho Hey guys, I hope you could help me out understanding the problem. From what I understand, the problem is to find the midpoint of a line (or two points). However, I do not understand the output of the given example. How can (2, 2) be the midpoint of (0, 0) and (1, 1)? Shouldn't it be (1/2, 1/2). I guess missing something, but I can't figure it out.
abhiranjan Nope, you have to find another point such that (1, 1) lies in the middle of (0,0) and the new point.
mrcoelho After your comment I read it again and this time I got it. I don't know how I've could have missed that. I guess I was tired. Thanks mate!
sakshamnsit i did the same but not geeting the answer for testcase2
yadavsunil00001 dont forget to put endl after printing both number.
rajan_kr [deleted]kunnu120 Thanks abhiranjan, the question's not straightforward.
sanjaychandak it is just different as you interpret it. the language is quite wrong. in this question you have to find the 'r'. mean to say q is the symmetrical point of p & r.
selftaught91 All the three points should lie on the same line for symmetricity to happen.
sahil7459 hey you doesn't need to find the mid point but you can consider q as a mid point like p=(1,1) q=(5,5) and then you need to find the r with will be at equal distance from the q like as p is, and it would be 180 i.e. p,q,r must pe point of the same strainght line so r=(q1+(q1-p1), q2+(q2-p2))
i.e. r=(5+(5-1),5+(5-1))
abolisetti python is bless
for i in range(int(input())): x1,y1,x2,y2 = (map(int, input().split())) print(x1+(x2-x1)*2, y1+(y2-y1)*2)
coarang12 here is solution for c#
int[] values = new int[2]; int rx = qx - px + qx, ry = qy - py + qy; values[0] = rx; values[1] = ry; return values;
160216735012ECE this is complete code in c. logic: arithematic progression code:
#include<stdio.h> int main(void) { int n; scanf("%d",&n); int a[n][n],m[n]; for(int i=0;i<n;i++) { scanf("%d%d%d%d", &a[i][1], &a[i][2], &a[i][3],&a[i][4]); } for(int i=0;i<n;i++) { printf("%d %d\n",a[i][1]+2*(a[i][3]-a[i][1]),a[i][2]+2*(a[i][4]-a[i][2])); } return 0; }
160216735034ECE GOOD. it's working :-)
160216735012ECE thank you buddy
dennindalke Minimalistic solution in Swift 4:
for _ in 1...Int(readLine()!)! { let line = readLine()!.split(separator:" ").map({Int($0)!}) print("\(2 * line[2] - line[0]) \(2 * line[3] - line[1])") }
dennindalke Minimalistic C# version:
using System.Linq; using static System.Console; class Solution { static void Main(string[] args) { for (int t = int.Parse(ReadLine()); t > 0; t--) { var ln = ReadLine().Split(' ').Select(int.Parse).ToArray(); WriteLine($"{2 * ln[2] - ln[0]} {2 * ln[3] - ln[1]}"); } } }
farti C#
int[] findPoint = new int[2]; findPoint[0]=(2*qx-px); findPoint[1]=(2*qy-py); return findPoint;
edlaierii What the what??? I cannot use the C# Point class from either System.Drawing or System.Windows. Well shucks!
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