Find X and Y intercepts of a line passing through the given points Last Updated : 06 Jan, 2024 Comments Improve Suggest changes Like Article Like Report Given two points on a 2D plane, the task is to find the x - intercept and the y - intercept of a line passing through the given points.Examples: Input: points[][] = {{5, 2}, {2, 7}} Output: 6.2 10.333333333333334Input: points[][] = {{3, 2}, {2, 4}} Output: 4.0 8.0 Approach: Find the slope using the given points.Put the value of the slope in the expression of the line i.e. y = mx + c.Now find the value of c using the values of any of the given points in the equation y = mx + cTo find the x-intercept, put y = 0 in y = mx + c.To find the y-intercept, put x = 0 in y = mx + c.Below is the implementation of the above approach: C++ // C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to find the X and Y intercepts // of the line passing through // the given points void getXandYintercept(int P[], int Q[]) { int a = P[1] - Q[1]; int b = P[0] - Q[0]; // if line is parallel to y axis if (b == 0) { cout << P[0] << endl; // x - intercept will be p[0] cout << "infinity"; // y - intercept will be infinity return; } // if line is parallel to x axis if (a == 0) { cout << "infinity"; // x - intercept will be infinity cout << P[1] << endl; // y - intercept will be p[1] return; } // Slope of the line double m = a / (b * 1.0); // y = mx + c in where c is unknown // Use any of the given point to find c int x = P[0]; int y = P[1]; double c = y - m * x; // For finding the x-intercept put y = 0 y = 0; double r = (y - c) / (m * 1.0); cout << r << endl; // For finding the y-intercept put x = 0 x = 0; y = m * x + c; printf("%.8f", c); } // Driver code int main() { int p1[] = { 5, 2 }; int p2[] = { 2, 7 }; getXandYintercept(p1, p2); return 0; } // This code is contributed by Mohit Kumar Java // Java implementation of the approach class GFG { // Function to find the X and Y intercepts // of the line passing through // the given points static void getXandYintercept(int P[], int Q[]) { int a = P[1] - Q[1]; int b = P[0] - Q[0]; // if line is parallel to y axis if (b == 0) { // x - intercept will be p[0] System.out.println(P[0]); // y - intercept will be infinity System.out.println("infinity"); return; } // if line is parallel to x axis if (a == 0) { // x - intercept will be infinity System.out.println("infinity"); // y - intercept will be p[1] System.out.println(P[1]); return; } // Slope of the line double m = a / (b * 1.0); // y = mx + c in where c is unknown // Use any of the given point to find c int x = P[0]; int y = P[1]; double c = y - m * x; // For finding the x-intercept put y = 0 y = 0; double r = (y - c) / (m * 1.0); System.out.println(r); // For finding the y-intercept put x = 0 x = 0; y = (int)(m * x + c); System.out.print(c); } // Driver code public static void main(String[] args) { int p1[] = { 5, 2 }; int p2[] = { 2, 7 }; getXandYintercept(p1, p2); } } // This code is contributed by kanugargng Python3 # Python3 implementation of the approach # Function to find the X and Y intercepts # of the line passing through # the given points def getXandYintercept(P, Q): a = P[1] - Q[1] b = P[0] - Q[0] # if line is parallel to y axis if b == 0: print(P[0]) # x - intercept will be p[0] print("infinity") # y - intercept will be infinity return # if line is parallel to x axis if a == 0: print("infinity") # x - intercept will be infinity print(P[1]) # y - intercept will be p[1] return # Slope of the line m = a / b # y = mx + c in where c is unknown # Use any of the given point to find c x = P[0] y = P[1] c = y-m * x # For finding the x-intercept put y = 0 y = 0 x =(y-c)/m print(x) # For finding the y-intercept put x = 0 x = 0 y = m * x + c print(y) # Driver code p1 = [5, 2] p2 = [7, 2] getXandYintercept(p1, p2) C# // C# implementation of the approach using System; class GFG { // Function to find the X and Y intercepts // of the line passing through // the given points static void getXandYintercept(int[] P, int[] Q) { int a = P[1] - Q[1]; int b = P[0] - Q[0]; // if line is parallel to y axis if (b == 0) { Console.WriteLine(P[0]); // x - intercept will be p[0] Console.WriteLine("infinity"); // y - intercept will be infinity return; } // if line is parallel to x axis if (a == 0) { Console.WriteLine("infinity"); // x - intercept will be infinity Console.WriteLine(P[1]); // y - intercept will be p[1] return; } // Slope of the line double m = a / (b * 1.0); // y = mx + c in where c is unknown // Use any of the given point to find c int x = P[0]; int y = P[1]; double c = y - m * x; // For finding the x-intercept put y = 0 y = 0; double r = (y - c) / (m * 1.0); Console.WriteLine(r); // For finding the y-intercept put x = 0 x = 0; y = (int)(m * x + c); Console.WriteLine(c); } // Driver code public static void Main() { int[] p1 = { 5, 2 }; int[] p2 = { 2, 7 }; getXandYintercept(p1, p2); } } // This code is contributed by AnkitRai01 JavaScript <script> // Javascript implementation of the approach // Function to find the X and Y intercepts // of the line passing through // the given points function getXandYintercept(P, Q) { let a = P[1] - Q[1]; let b = P[0] - Q[0]; // if line is parallel to y axis if (b == 0) { document.write(P[0] + "</br>"); // x - intercept will be p[0] document.write("infinity" + "</br>"); // y - intercept will be infinity return; } // if line is parallel to x axis if (a == 0) { document.write("infinity" + "</br>"); // x - intercept will be infinity document.write(P[1] + "</br>"); // y - intercept will be p[1] return; } // Slope of the line let m = a / (b * 1.0); // y = mx + c in where c is unknown // Use any of the given point to find c let x = P[0]; let y = P[1]; let c = y - m * x; // For finding the x-intercept put y = 0 y = 0; let r = (y - c) / (m * 1.0); document.write(r + "</br>"); // For finding the y-intercept put x = 0 x = 0; y = parseInt(m * x + c, 10); document.write(c.toFixed(11) + "</br>"); } let p1 = [ 5, 2 ]; let p2 = [ 2, 7 ]; getXandYintercept(p1, p2); </script> Output6.2 10.33333333333 Time Complexity: O(1)Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Find X and Y intercepts of a line passing through the given points A Atul_kumar_Shrivastava Follow Improve Article Tags : Algorithms Mathematical DSA Geometric-Lines Practice Tags : AlgorithmsMathematical Similar Reads DSA Tutorial - Learn Data Structures and Algorithms DSA (Data Structures and Algorithms) is the study of organizing data efficiently using data structures like arrays, stacks, and trees, paired with step-by-step procedures (or algorithms) to solve problems effectively. Data structures manage how data is stored and accessed, while algorithms focus on 7 min read Quick Sort QuickSort is a sorting algorithm based on the Divide and Conquer that picks an element as a pivot and partitions the given array around the picked pivot by placing the pivot in its correct position in the sorted array. It works on the principle of divide and conquer, breaking down the problem into s 12 min read Merge Sort - Data Structure and Algorithms Tutorials Merge sort is a popular sorting algorithm known for its efficiency and stability. It follows the divide-and-conquer approach. It works by recursively dividing the input array into two halves, recursively sorting the two halves and finally merging them back together to obtain the sorted array. Merge 14 min read Bubble Sort Algorithm Bubble Sort is the simplest sorting algorithm that works by repeatedly swapping the adjacent elements if they are in the wrong order. This algorithm is not suitable for large data sets as its average and worst-case time complexity are quite high.We sort the array using multiple passes. After the fir 8 min read Data Structures Tutorial Data structures are the fundamental building blocks of computer programming. They define how data is organized, stored, and manipulated within a program. Understanding data structures is very important for developing efficient and effective algorithms. What is Data Structure?A data structure is a st 2 min read Breadth First Search or BFS for a Graph Given a undirected graph represented by an adjacency list adj, where each adj[i] represents the list of vertices connected to vertex i. Perform a Breadth First Search (BFS) traversal starting from vertex 0, visiting vertices from left to right according to the adjacency list, and return a list conta 15+ min read Binary Search Algorithm - Iterative and Recursive Implementation Binary Search Algorithm is a searching algorithm used in a sorted array by repeatedly dividing the search interval in half. The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(log N). Binary Search AlgorithmConditions to apply Binary Searc 15 min read Insertion Sort Algorithm Insertion sort is a simple sorting algorithm that works by iteratively inserting each element of an unsorted list into its correct position in a sorted portion of the list. It is like sorting playing cards in your hands. You split the cards into two groups: the sorted cards and the unsorted cards. T 9 min read Dijkstra's Algorithm to find Shortest Paths from a Source to all Given a weighted undirected graph represented as an edge list and a source vertex src, find the shortest path distances from the source vertex to all other vertices in the graph. The graph contains V vertices, numbered from 0 to V - 1.Note: The given graph does not contain any negative edge. Example 12 min read Selection Sort Selection Sort is a comparison-based sorting algorithm. It sorts an array by repeatedly selecting the smallest (or largest) element from the unsorted portion and swapping it with the first unsorted element. This process continues until the entire array is sorted.First we find the smallest element an 8 min read Like