Program to find line passing through 2 Points (original) (raw)

Last Updated : 23 Jul, 2025

Given two points P and Q in the coordinate plane, find the equation of the line passing through both points.
This kind of conversion is very useful in many geometric algorithms like intersection of lines, finding the circumcenter of a triangle, finding the incenter of a triangle and many more...

Examples:

Input : P(3, 2) Q(2, 6) Output : 4x + 1y = 14

Input : P(0, 1) Q(2, 4) Output : 3x + -2y = -2

Let the given two points be P(x1, y1) and Q(x2, y2). Now, we find the equation of line formed by these points.
Any line can be represented as,
ax + by = c
Let the two points satisfy the given line. So, we have,
ax1 + by1 = c
ax2 + by2 = c

We can set the following values so that all the equations hold true,

a = y2 - y1 b = x1 - x2 c = ax1 + by1

These can be derived by first getting the slope directly and then finding the intercept of the line. OR these can also be derived cleverly by a simple observation as under:

Derivation :

ax1 + by1 = c ...(i) ax2 + by2 = c ...(ii) Equating (i) and (ii), ax1 + by1 = ax2 + by2 => a(x1 - x2) = b(y2 - y1) Thus, for equating LHS and RHS, we can simply have, a = (y2 - y1) AND b = (x1 - x2) so that we have, (y2 - y1)(x1 - x2) = (x1 - x2)(y2 - y1) AND Putting these values in (i), we get, c = ax1 + by1

Thus, we now have the values of a, b, and c which means that we have the line in the coordinate plane.

Try It Yourself

C++ `

// C++ Implementation to find the line passing // through two points #include using namespace std;

// This pair is used to store the X and Y // coordinate of a point respectively #define pdd pair<double, double>

// Function to find the line given two points void lineFromPoints(pdd P, pdd Q) { double a = Q.second - P.second; double b = P.first - Q.first; double c = a * (P.first) + b * (P.second);

if (b < 0) {
    cout << "The line passing through points P and Q "
            "is: "
         << a << "x - " << b << "y = " << c << endl;
}
else {
    cout << "The line passing through points P and Q "
            "is: "
         << a << "x + " << b << "y = " << c << endl;
}

}

// Driver code int main() { pdd P = make_pair(3, 2); pdd Q = make_pair(2, 6); lineFromPoints(P, Q); return 0; }

Java

// Java Implementation to find the line passing // through two points import java.io.*; public class GFG {

// This pair is used to store the X and Y
// coordinate of a point respectively
static class Pair {
    int first, second;

    public Pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function to find the line given two points
static void lineFromPoints(Pair P, Pair Q)
{
    int a = Q.second - P.second;
    int b = P.first - Q.first;
    int c = a * (P.first) + b * (P.second);

    if (b < 0) {
        System.out.println(
            "The line passing through points P and Q is: "
            + a + "x - " + b + "y = " + c);
    }
    else {
        System.out.println(
            "The line passing through points P and Q is: "
            + a + "x + " + b + "y = " + c);
    }
}

// Driver code
public static void main(String[] args)
{
    Pair P = new Pair(3, 2);
    Pair Q = new Pair(2, 6);
    lineFromPoints(P, Q);
}

}

// This code is contributed by Princi Singh

Python3

Python3 Implementation to find the line passing

through two points

This pair is used to store the X and Y

coordinate of a point respectively

define pdd pair<double, double>

Function to find the line given two points

def lineFromPoints(P, Q):

a = Q[1] - P[1]
b = P[0] - Q[0]
c = a*(P[0]) + b*(P[1])

if(b < 0):
    print("The line passing through points P and Q is:",
          a, "x - ", b, "y = ", c, "\n")
else:
    print("The line passing through points P and Q is: ",
          a, "x + ", b, "y = ", c, "\n")

Driver code

if name == 'main': P = [3, 2] Q = [2, 6] lineFromPoints(P, Q)

This code is contributed by ash264

C#

// C# Implementation to find the line passing // through two points using System;

class GFG {

// This pair is used to store the X and Y
// coordinate of a point respectively
public class Pair {
    public int first, second;

    public Pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function to find the line given two points
static void lineFromPoints(Pair P, Pair Q)
{
    int a = Q.second - P.second;
    int b = P.first - Q.first;
    int c = a * (P.first) + b * (P.second);

    if (b < 0) {
        Console.WriteLine(
            "The line passing through points P and Q is: "
            + a + "x - " + b + "y = " + c);
    }
    else {
        Console.WriteLine(
            "The line passing through points P and Q is: "
            + a + "x + " + b + "y = " + c);
    }
}

// Driver code
public static void Main(String[] args)
{
    Pair P = new Pair(3, 2);
    Pair Q = new Pair(2, 6);
    lineFromPoints(P, Q);
}

}

// This code has been contributed by 29AjayKumar

JavaScript

`

Output

The line passing through points P and Q is: 4x + 1y = 14

Time Complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.