# Find foot of perpendicular from a point in 2D plane to a Line in C++

Consider we have a point P in 2D plane and equation of a line, the task is to find the foot of the perpendicular from P to the line.

The equation of the straight line is ax + by + c = 0. Equation of line passing through P and perpendicular to line. Equation of line passing through P and Q will be ay – bx + d = 0. Also P(x1, y1), and Q(x2, y2), so we put coordinate of P on the equation.

ay 1−bx 1+d=0, so d=bx1−ay 1

Also Q is the intersection of the given line and the line passing through P and Q, so we will find solution for these two equations.

ax+by+c=0,∧ay−bx+(bx1−ay 1)=0

As a, b, c, d all are known, we can find using this formula −

$$\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{f-(ax_{1}+by_{1}+c)}{a^{2}+b^{2}}$$

## Example

Live Demo

#include<iostream>
using namespace std;
void getFootCoordinate(double a, double b, double c, double x1, double y1) {
double p = -1 * (a * x1 + b * y1 + c) / (a * a + b * b);
double x = p * a + x1;
double y = p * b + y1;
cout << "(" << x << ", " << y <<")";
}
int main() {
double a = 0.0;
double b = 1.0;
double c = -2;
double x1 = 3.0;
double y1 = 3.0;
cout << "The coordinate is: ";
getFootCoordinate(a, b, c, x1, y1);
}

## Output

The coordinate is: (3, 2)