# Contest Name:

**CODICTION 2020**

# PROBLEM LINK:

* Author:* Srijan Agrawal

*Mradul Rathore, Nehal Jain, Saniya Agrawal*

**Tester:***Murtaza Ali*

**Editorialist:**# DIFFICULTY:

CAKEWALK.

# PREREQUISITES:

Geometry.

# PROBLEM:

Find a point on x-axis such that the reflection of (x_{2}, y_{2}) can be seen from the point (x_{1}, y_{1}).

# QUICK EXPLANATION:

We have to place a mirror on (x, 0) such that the reflection of (x_{2}, y_{2}) in the mirror on (x, 0) can be seen from the point (x_{1}, y_{1}).

# EXPLANATION:

The Law of Reflection states that the angle of incidence is equal to the angle of reflection. Let the ray passing through (x_{2}, y_{2}) and (x, 0) be the incident line and the ray from (x_{1}, y_{1}) and (x, 0) be the reflected line.

Let,

Angle of incidence = Angle of reflection = \theta

Slope of line of incidence = tan(\frac{\pi}{2} - \theta) = cot(\theta)

Slope of line of reflection = tan(\frac{\pi}{2} + \theta) = - cot(\theta)

Slope of line of incidence in two point form = \frac{y_{2} - 0}{x_{2} - x} = \frac{y_{2}}{x_{2} - x}

Slope of line of reflection in two point form = \frac{y_{1} - 0}{x_{1} - x} = \frac{y_{1}}{x_{1} - x}

By above equations, we can get,

\frac{y_{2}}{x_{2} - x} = - \frac{y_{1}}{x_{1} - x}

x = \frac{x_{2}*y_{1} + x_{1}*y_{2}}{y_{1} + y_{2}}

# COMPLEXITY:

**Time complexity:** O(1) per test.

# SOLUTIONS:

## Setter's Solution

```
#include<bits/stdc++.h>
using namespace std;
int main()
{
int T;
cin>>T;
while(T--){
long long int x1,y1,x2,y2;
cin>>x1>>y1>>x2>>y2;
double n=(x2*y1+y2*x1);
double ans=n/(y1+y2);
cout << fixed << setprecision(2)<<ans<<endl;
}
return 0;
}
```

# ALTERNATE SOLUTION:

We can also solve this problem using **Similarity of Triangles**. If you want detailed explanation for this approach please let us know.

## Alternate Solution by Tester

```
#include <iostream>
using namespace std;
int main() {
int t;
cin>>t;
while(t--){
int x1, y1, x2, y2;
cin>>x1>>y1;
cin>>x2>>y2;
float r = (float) abs(x1 - x2) / (y1 + y2);
if(x1 < x2){
r *= y1;
r += x1;
}
else{
r *= y2;
r += x2;
}
printf("%0.2f\n", r);
}
return 0;
}
```

Feel free to share your approach. If you have any queries, they are always welcome.