Drawing line perpendicular to a given line

问题:

I have start and end coordinate of a line. I want to drawn another line sticking at the end of this this such that they will be perpendicular to each other.

I am trying to do this using the normal geometry. Is there any high-level API there in MFC for the same.

Thanks

回答1:

If (dx,dy) are the differences in the x and y coodinates of the given line, you can make another line perpendicular by contriving for the differences in its coordinates to be (-dy, dx). You can scale that by any factor (-c*dy, c*dx) to change its length.

回答2:

You have an existing line (x1, y1) to (x2, y2). The perpendicular line is (a1, b1) to (a2, b2), and centered on (x2, y2).

xdif = x2 - x1
ydif = y2 - y1
a1 = x2 - ydif / 2
b1 = y2 + xdif / 2
a2 = x2 + ydif / 2
b2 = y2 - xdif / 2

I think that works... I tested it for a few lines.

So if you have a line going from (1,1) to (5,3), the perpendicular line would be (5 - 2/2, 3+4/2) to (5 + 2/2, 3 - 4/2) or (4,5) to (6, 1).

回答3:

You could use SetWorldTransform function from Win32 GDI API.

Sample code is here.

回答4:

Let me add some c++ code based on kbelder answer. It make one vertex by origin point (x1,y1) and another vertex (x2,y2)

float GetDistance(float x1, float y1, float x2, float y2)
{
 float cx = x2 - x1;
 float cy = y2 - y1;
 float flen = sqrtf((float)(cx*cx + cy*cy));
 return flen;
}

void GetAxePoint(double x1, double y1, double x2, double y2, double& x3, double& y3, double vec_len, bool second_is_y)
{
 double xdif = x2 - x1;
 double ydif = y2 - y1;

 if(second_is_y)
 {
  x3 = x1 - ydif;
  y3 = y1 + xdif;
 }
 else
 {
  x3 = x1 + ydif;
  y3 = y1 - xdif;
 }

 double vec3_len = GetDistance(x3, y3, x1, y1);

 x3 = (x3-x1)/vec3_len;
 y3 = (y3-y1)/vec3_len;

 x3 = x1 + x3*vec_len;
 y3 = y1 + y3*vec_len;
}