objective-c implements the distance from a point to a straight line and the intersection point with the vertical foot

Problem description

The distance from a point to a straight line or line segment

Implementation ideas

Suppose there is a point coordinate P (x0, y0), a line segment AB, A coordinate (x1, y1), and B coordinate (x2, y2), find the distance d from point P to line segment AB or the straight line where it is located, and the vertical foot C (x, y) of point P on the straight line.

This requires reviewing high school mathematics knowledge.

First, you need to convert the coordinates of points A and B into the general formula of the linear equation Ax+By+C = 0, and the process will not be deduced.

Parameter calculation:

A=y2-y1；

B=x1-x2；

C=x2*y1-x1*y2;

1. The distance formula from a point to a straight line:

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

2. Calculation formula for C (x, y) of the hang 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 );

Program implementation:

-(CGPoint)pedalPoint: (CGPoint)p1 : (CGPoint )p2: (CGPoint)x0{

float A=;
float B=;
float C=**;

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

//Distance from point to linefloat d=(A*+B*+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(@”Drooping feet======x=%f,y=%f”,x,y);
return ptCross;
}

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