ron@clarity.Princeton.EDU (Ronald Beekelaar) (11/18/89)
Hi everyone,
I know this is a silly question. But I need the answer and could think of
another way of getting it. Actually it is a trig-problem I ran into, while
building a graphical interface. Here it is:
" What is the shortest distance between coordinate (a3, b3) and the line
through coordinates (a1, b1) and (a2, b2)? "
The answer should be something like dis = < some formula >, but I couldn't
figure out which one it was.
Thanks for helping me
--
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ron
------ajaym@cbnewsc.ATT.COM (alton.jay.mitchell) (11/18/89)
> " What is the shortest distance between coordinate (a3, b3) and the line > through coordinates (a1, b1) and (a2, b2)? " I didn't take the time to compute a formula, but it seems somewhat obvious that the answer can be presented in 3 parts: (a1,b1)------------------(a2,b2) If (a3,b3) is "within" the area bounded by the perpendicular lines through (a1,b1) and (a2,b2), then the shortest distance is the perpendicular line from (a3,b3) to the line shown. Otherwise, the shortest distance is the line to one of the 2 points, depending on which "side" of the above mentioned area (a3,b3) exists on.
dopey%looney@Sun.COM (Can't ya tell by the name) (11/21/89)
In article <4881@cbnewsc.ATT.COM> ajaym@cbnewsc.ATT.COM (alton.jay.mitchell) writes: >> " What is the shortest distance between coordinate (a3, b3) and the line >> through coordinates (a1, b1) and (a2, b2)? " > >I didn't take the time to compute a formula, but it seems somewhat >obvious that the answer can be presented in 3 parts: > > (a1,b1)------------------(a2,b2) > >If (a3,b3) is "within" the area bounded by the perpendicular >lines through (a1,b1) and (a2,b2), then the shortest distance >is the perpendicular line from (a3,b3) to the line shown. > >Otherwise, the shortest distance is the line to one of the >2 points, depending on which "side" of the above mentioned >area (a3,b3) exists on. In nitty gritty detail (this works but may not be the BEST way): I stole this from some code I had written before and checked it over but I have not run THIS version but I hope it helps. It only computes the SQUARE of the distance because I was only looking for relative distances and finding the square root was to much over head. /* This routine computes the SQUARE of the * distance from point x,y to line l */ struct line /* format of line overlay */ { short label; /* indecates type */ long x1; /* start x */ long y1; /* start y */ long x2; /* end x */ long y2; /* end y */ }; #define SIGN(x) (((x) >= 0)? 1: -1) double hypot(); double point_line(x, y, l) int x; int y; struct line *l; { double a; /* slop of the line */ double b; /* -1 in double form */ double c; /* y intercept */ double d0; /* distance from x,y to "infinite" line */ double d1; /* length of line */ double d2; /* worst case of point to line */ double d3; /* distance from x,y to line endpoint 1 */ double d4; /* distance from x,y to line endpoint 2 */ double d; /* distance from x,y to line */ if((l->x2 - l->x1) == 0) { d0 = MIN(ABS(l->y2 - y), ABS(l->y1 - y)); } else { a = (double)(l->y2 - l->y1)/(double)(l->x2 - l->x1); b = -1; c = l->y1 - a*l->x1; d0 = (double)(a*x + b*y + c)/(double)(-1*SIGN(c)*hypot(a,b)); d0 = ABS(d0); } d1 = hypot((double)ABS(l->x2 - l->x1), (double)ABS(l->y2 - l->y1)); d2 = hypot(d0, d1); d3 = hypot((double)ABS(l->x1 - x), (double)ABS(l->y1 - y)); d4 = hypot((double)ABS(l->x2 - x), (double)ABS(l->y2 - y)); d = d0; if (d3 > d2 || d4 > d2) { d = MIN(d3,d4); } return(ABS(d)); }