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+## LOJ-1292
+# Laser Shot
+
+### Simplified Version
+You have points on a flat surface (xy plane). You want to draw a line in a way that the most points possible lie on that line. What's the highest number of points that can be on one line?
+
+### Tags
+Geometry, Slope
+
+# Editorial
+
+## Observation
+1. If two lines have the same slope, they are parallel.
+
+2. If a point lies on two parallel lines, those lines are actually the same line.
+
+![image of a slope](slope.webp)
+
+First, for each of the n points, pick one point and determine the slope between that point and all the other points.
+
+The most common slope value gives us the most points on a line drawn from a certain point.
+
+By repeating this process for every point, the highest count of shared slopes becomes the solution.
+
+### How to Calculate Slope:
+
+If two points are (x1, y1) and (x2, y2) then slope of those two point is,
+
+Let, a = x2 - x1, b = y2 - y1 , g = gcd(a, b).
+
+slope = (a/gcd)/(b/gcd)
+
+# Complexity Analysis
+
+## Time Complexity 
+O(n<sup>2</sup>logn)
+
+## Memory Complexity
+O(n)
+
+# Program Code
+
+```c++
+#include<bits/stdc++.h>
+using namespace std;
+int cas = 0;
+pair<int,int> calculate_slope(vector<int> point1, vector<int>point2)
+{
+    int dx = point1[0] - point2[0];
+    int dy = point1[1] - point2[1];
+    int gcd = __gcd(dx, dy);
+    
+    dx /= gcd;
+    dy /= gcd;
+    return {dx,dy};
+}
+void solve()
+{
+    int n;
+    cin >> n;
+    vector<vector<int>>points;
+    for(int i = 0; i < n; i++){
+        int x,y;
+        cin >> x >> y;
+        points.push_back({x,y});
+    }
+
+    map<pair<int,int>,int>slope_count;
+    int max_point = 0;
+    
+    for(int i = 0; i < points.size(); i++){
+        slope_count.clear();
+        for(int j = i+1; j < points.size(); j++){
+            //calculate the slope.
+            pair<int, int>slope = calculate_slope(points[i], points[j]);
+            slope_count[slope]++;
+        }
+        // Finding the maximum number of points in a single line
+        for(auto it : slope_count){
+            max_point = max(max_point, it.second);
+        }
+    }
+        
+    cout << "Case " << ++cas << ": " <<  max_point+1 << "\n";
+}
+int main()
+{
+    ios_base :: sync_with_stdio(0);
+    cin.tie(0);
+    int t;
+    cin >> t;
+    while(t--)solve();
+
+    return 0;
+}
+```
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