HP Prime: Minimum Distance Between a Point and a Line
Introduction
We have a line in the form of y = m * x + b, where m is the slope of the line and b is the y-intercept of the line, and a separate point (px, py). The task is to find the minimum distance, or the shortest distance, between the point and the line. The separate point is not required to be on the line. The line and point are in two-dimensional space.
If the point (px, py) is not on the line, then theoretically, there are an infinite amount of distances between the point and the line. However, to get the shortest distance, draw a path that is “directly straight” to the line. This is achieved by choosing a line that connects the (px, py) that is a line that is orthogonal (perpendicular) to the line y = m * x + b.
The line drawn is of the form y = -1/m * x + b1. The slope of the orthogonal line is -1/m. Assuming that m ≠ 0, the y-intercept of the orthogonal line is b1 = y1 + x1 / m.
The next step is to find where the two lines intersect, which is done by solving the following system for x and y:
y = m * x + b
y = -m / n + b1
Label the intersection point (x1, y1). The minimum distance will be calculated as follows:
dist = √( (x1 – px)^2 + (y1 – py)^2 ) = abs( (x1 – px) + (y1 – py)*i)
If m = 0, the line is in the form of y = b. The orthogonal line is x = px, and the distance is simply abs( (y1 – py)*i ).
HP Prime Code: PTLNDIST
EXPORT PTLNDIST()
BEGIN
// 2024-07-21 EWS
// radian
HAngle:=0;
LOCAL px,py,m,b;
INPUT({m,b,px,py},
"Point-Line Distance (px, py), y=mx+b",
{"m:","b:","px:","py:"},
{"m: slope"," b: y-intercept",
"point x","point y"});
LOCAL y0;
y0:=m*px+b;
LOCAL b1,mt,x1,y1,dist,str;
IF m≠0 THEN
b1:=py+px/m;
mt:=[[−m,1],[1/m,1]]^-1*[[b],[b1]];
x1:=mt[1,1];
y1:=mt[2,1];
dist:=ABS((x1-px)+(y1-py)*√(-1));
ELSE
x1:=px;
y1:=b;
dist:=ABS((y1-py)*√(-1));
END;
// print results
PRINT();
PRINT("Results:");
PRINT("Intersect point:");
PRINT("x: "+STRING(x1));
PRINT("y: "+STRING(y1));
PRINT("");
IF m≠0 THEN
str:="Y="+STRING(-1/m)+"*X+"+STRING(b1);
ELSE
str:="X="+STRING(x1);
END;
PRINT("Orthogonal Line:");
PRINT(str);
PRINT("");
PRINT("Minimum Distance:");
PRINT(dist);
RETURN {x1,y1,str,dist};
END;
Note:
√(-1) represents the imaginary number ⅈ ( [ Shift ], [ 2 ] ).
Inputs:
* The slope of the y-intercept of the line y = m * x + b (no vertical lines, but m can be zero)
* The point (px, py)
Outputs:
* The line that runs through point (px, yx) that is orthogonal to y = m * x + b. The slope and y-intercept of the orthogonal line, which the line will be stated in a string
* The intersection point of the two lines.
* The distance between (px, yx) and the intersection point. (dist)
Examples
Example 1:
Inputs: Line: y = 5 x – 2, Point: (-1, -5)
m = 5
b = -2
px = -1
py = -5
Results:
Intersect point:
x = -0.615384615386
y = -5.07692307692
Orthogonal Line:
Y = -0.2 * X – 5.2
Minimum distance:
0.392232270274
Example 2:
Inputs: Line: y = 6, Point: (3, -9)
m = 0
b = 6
px = 3
py = -9
Results:
Intersect point:
x = 0.764705882353
y = 4.05882352941
Orthogonal Line:
X = 3
Minimum distance:
15
Example 3:
Inputs: Line: y = 4 x + 1, Point: (5, 3)
m = 4
b = 1
px = 5
py = 3
Results:
Intersect point:
x = 0.764705882353
y = 4.05882352941
Orthogonal Line:
Y = -0.25 * X + 4.25
Minimum distance:
4.36564125066
Source
Tremblay, Christopher. Mathematics for Game Developers. Thomson Course Technology. Boston, MA. 2004. ISBN 1-59200-038-X.
Eddie
All original content copyright, © 2011-2024. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.
