objective c Distance from point to line and intersection with vertical foot

Problem description

The distance from a point to a line or segment

Realization thought

Assuming that there are 1 point coordinates P (x0, y0), 1 line segment AB, A coordinates (x1, y1) and B coordinates (x2, y2), find the distance d from P point to AB line segment or straight line, and the vertical foot C (x, y) of P point on straight line.

This requires a review of mathematics knowledge in high school.

Firstly, the coordinates of A and B need to be transformed into the general formula Ax+By+C = 0, so the process will not be deduced.

Parameter calculation:

A=y2-y1;

B=x1-x2;

C=x2*y1-x1*y2;

1. Distance formula from point to line:

d= ( Ax0 + By0 + C ) / sqrt ( A*A + B*B );

2. Calculation formula of C (x, y) of vertical foot:

x = ( B*B*x0 - A*B*y0 - A*C ) / ( A*A + B*B );

y = (-A*B*x0+A*A*y0-B*C)/(A*A + B*B);

Implementation of the program:

// Vertical foot intersection 
-(CGPoint)pedalPoint: (CGPoint)p1 : (CGPoint )p2: (CGPoint)x0{

float A=p2.y-p1.y;
float B=p1.x-p2.x;
float C=p2.x*p1.y-p1.x*p2.y;

float x=(B*B*x0.x-A*B*x0.y-A*C)/(A*A+B*B);
float y=(-A*B*x0.x+A*A*x0.y-B*C)/(A*A+B*B);

// Point-to-line distance 
float d=(A*x0.x+B*x0.y+C)/sqrt(A*A+B*B);

CGPoint ptCross=ccp(x,y);
NSLog(@ " d======%f " ,d);
NSLog(@ " A=======%f,B=======%f,C=======%f " ,A,B,C);
NSLog(@ "Hanging your feet ======x=%f,y=%f " ,x,y);
return ptCross;
}

Summarize