指向C中的直线特定距离

Question

如何在直线上找到距离给定点特定距离的点.我用C编写这段代码,但我得不到正确的答案.难道你们有人指导我做错了什么.

我得到x1,y1,x2,y2值和剩下的距离很好.使用这些我可以找到斜率m和y截距也很好.现在,我需要在连接这两个点的直线上找到点,该点距点x1,y1 10个单位.我好像在这里出错了.这是我写的代码.

int x1 = node[n].currentCoordinates.xCoordinate;
int y1 = node[n].currentCoordinates.yCoordinate;
int x2 = node[n].destinationLocationCoordinates.xCoordinate;
int y2 = node[n].destinationLocationCoordinates.yCoordinate;

int distanceleft = (y2 - y1) * (y2 - y1) + (x2 - x1) * (x2 - x1);
distanceleft = sqrt(distanceleft);
printf("Distance left to cover is %d\n",distanceleft);
int m = (y2 - y1)/(x2 - x1); // slope.
int b = y1 - m * x1; //y-intercept


//find point on the line that is 10 units away from
//current coordinates on equation y = mx + b.
if(x2 > x1)
{
     printf("x2 is greater than x1\n");
     int tempx = 0;
     int tempy = 0;
     for(tempx = x1; tempx <= x2; tempx++)
     {
          tempy = y1 + (y2 - y1) * (tempx - x1)/(x2 - x1);
          printf("tempx = %d, tempy = %d\n",tempx,tempy);
          int distanceofthispoint = (tempy - y1) * (tempy - y1) + (tempx - x1) * (tempx - x1);
          distanceofthispoint = sqrt((int)distanceofthispoint);
          if(distanceofthispoint >= 10)
          {
               //found new points.
               node[n].currentCoordinates.xCoordinate = tempx;
               node[n].currentCoordinates.yCoordinate = tempy;
               node[n].TimeAtCurrentCoordinate = clock;
               printf("Found the point at the matching distance\n");
               break;
          }
     }
}
else
{
     printf("x2 is lesser than x1\n");
     int tempx = 0;
     int tempy = 0;
     for(tempx = x1; tempx >= x2; tempx--)
     {
          tempy = y1 + (y2 - y1) * (tempx - x1)/(x2 - x1);
          printf("tempx = %d, tempy = %d\n",tempx,tempy);
          int distanceofthispoint = (tempy - y1) * (tempy - y1) + (tempx - x1) * (tempx - x1);
          distanceofthispoint = sqrt((int)distanceofthispoint);
          if(distanceofthispoint >= 10)
          {
               //found new points.
               node[n].currentCoordinates.xCoordinate = tempx;
               node[n].currentCoordinates.yCoordinate = tempy;
               node[n].TimeAtCurrentCoordinate = clock;
               printf("Found the point at the matching distance\n");
               break;
          }
     }
}
printf("at time %f, (%d,%d) are the coordinates of node %d\n",clock,node[n].currentCoordinates.xCoordinate,node[n].currentCoordinates.yCoordinate,n);

Answer 1

这是数学的方式,我没时间用C写东西.

你有一个点(x1,y1)和另一个点(x2,y2),当链接它给你一个段.

因此,你有一个方向向量v=(xv, yv),其中xv=x2-x1和yv=y2-y1.

现在,你需要将这个向量除以它的范数,得到一个新的向量:vector = v/sqrt(xv ² + yv ²).

现在,您只需添加到原点,向量乘以您想要点的距离:

位置=(x原点,y原点)+距离×矢量

我希望这有帮助!