问题
How can I draw a perpendicular on a line segment from a given point? My line segment is defined as (x1, y1), (x2, y2), If I draw a perpendicular from a point (x3,y3) and it meets to line on point (x4,y4). I want to find out this (x4,y4).
回答1:
I solved the equations for you:
k = ((y2-y1) * (x3-x1) - (x2-x1) * (y3-y1)) / ((y2-y1)^2 + (x2-x1)^2)
x4 = x3 - k * (y2-y1)
y4 = y3 + k * (x2-x1)
Where ^2 means squared
回答2:
From wiki:
In algebra, for any linear equation y=mx + b, the perpendiculars will all have a slope of (-1/m), the opposite reciprocal of the original slope. It is helpful to memorize the slogan "to find the slope of the perpendicular line, flip the fraction and change the sign." Recall that any whole number a is itself over one, and can be written as (a/1)
To find the perpendicular of a given line which also passes through a particular point (x, y), solve the equation y = (-1/m)x + b, substituting in the known values of m, x, and y to solve for b.
The slope of the line, m, through (x1, y1) and (x2, y2) is m = (y1 - y2) / (x1 - x2)
回答3:
I agree with peter.murray.rust, vectors make the solution clearer:
// first convert line to normalized unit vector
double dx = x2 - x1;
double dy = y2 - y1;
double mag = sqrt(dx*dx + dy*dy);
dx /= mag;
dy /= mag;
// translate the point and get the dot product
double lambda = (dx * (x3 - x1)) + (dy * (y3 - y1));
x4 = (dx * lambda) + x1;
y4 = (dy * lambda) + y1;
回答4:
You will often find that using vectors makes the solution clearer...
Here is a routine from my own library:
public class Line2 {
Real2 from;
Real2 to;
Vector2 vector;
Vector2 unitVector = null;
public Real2 getNearestPointOnLine(Real2 point) {
unitVector = to.subtract(from).getUnitVector();
Vector2 lp = new Vector2(point.subtract(this.from));
double lambda = unitVector.dotProduct(lp);
Real2 vv = unitVector.multiplyBy(lambda);
return from.plus(vv);
}
}
You will have to implement Real2 (a point) and Vector2 and dotProduct() but these should be simple:
The code then looks something like:
Point2 p1 = new Point2(x1, y1);
Point2 p2 = new Point2(x2, y2);
Point2 p3 = new Point2(x3, y3);
Line2 line = new Line2(p1, p2);
Point2 p4 = getNearestPointOnLine(p3);
The library (org.xmlcml.euclid) is at: http://sourceforge.net/projects/cml/
and there are unit tests which will exercise this method and show you how to use it.
@Test
public final void testGetNearestPointOnLine() {
Real2 p = l1112.getNearestPointOnLine(new Real2(0., 0.));
Real2Test.assertEquals("point", new Real2(0.4, -0.2), p, 0.0000001);
}
回答5:
You know both the point and the slope, so the equation for the new line is:
y-y3=m*(x-x3)
Since the line is perpendicular, the slope is the negative reciprocal. You now have two equations and can solve for their intersection.
y-y3=-(1/m)*(x-x3)
y-y1=m*(x-x1)
回答6:
Compute the slope of the line joining points (x1,y1) and (x2,y2) as m=(y2-y1)/(x2-x1)
Equation of the line joining (x1,y1) and (x2,y2) using point-slope form of line equation, would be y-y2 = m(x-x2)
Slope of the line joining (x3,y3) and (x4,y4) would be -(1/m)
Again, equation of the line joining (x3,y3) and (x4,y4) using point-slope form of line equation, would be y-y3 = -(1/m)(x-x3)
Solve these two line equations as you solve a linear equation in two variables and the values of x and y you get would be your (x4,y4)
I hope this helps.
cheers
回答7:
Find out the slopes for both the lines, say slopes are m1 and m2 then m1*m2=-1 is the condition for perpendicularity.
回答8:
Matlab function code for the following problem
function Pr=getSpPoint(Line,Point)
% getSpPoint(): find Perpendicular on a line segment from a given point
x1=Line(1,1);
y1=Line(1,2);
x2=Line(2,1);
y2=Line(2,1);
x3=Point(1,1);
y3=Point(1,2);
px = x2-x1;
py = y2-y1;
dAB = px*px + py*py;
u = ((x3 - x1) * px + (y3 - y1) * py) / dAB;
x = x1 + u * px;
y = y1 + u * py;
Pr=[x,y];
end
回答9:
Mathematica introduced the function RegionNearest[] in version 10, 2014. This function could be used to return an answer to this question:
{x4,y4} = RegionNearest[Line[{{x1,y1},{x2,y2}}],{x3,y3}]
回答10:
This is mostly a duplicate of Arnkrishn's answer. I just wanted to complete his section with a complete Mathematica code snippet:
m = (y2 - y1)/(x2 - x1)
eqn1 = y - y3 == -(1/m)*(x - x3)
eqn2 = y - y1 == m*(x - x1)
Solve[eqn1 && eqn2, {x, y}]
回答11:
This is a C# implementation of the accepted answer. It's also using ArcGis to return a MapPoint as that's what we're using for this project.
private MapPoint GenerateLinePoint(double startPointX, double startPointY, double endPointX, double endPointY, double pointX, double pointY)
{
double k = ((endPointY - startPointY) * (pointX - startPointX) - (endPointX - startPointX) * (pointY - startPointY)) / (Math.Pow(endPointY - startPointY, 2)
+ Math.Pow(endPointX - startPointX, 2));
double resultX = pointX - k * (endPointY - startPointY);
double resultY = pointY + k * (endPointX - startPointX);
return new MapPoint(resultX, resultY, 0, SpatialReferences.Wgs84);
}
Thanks to Ray as this worked perfectly for me. c#arcgis
回答12:
This is a vectorized Matlab function for finding pairwise projections of m points onto n line segments. Here xp and yp are m by 1 vectors holding coordinates of m different points, and x1, y1, x2 and y2 are n by 1 vectors holding coordinates of start and end points of n different line segments.
It returns m by n matrices, x and y, where x(i, j) and y(i, j) are coordinates of projection of i-th point onto j-th line.
The actual work is done in first few lines and the rest of the function runs a self-test demo, just in case where it is called with no parameters. It's relatively fast, I managed to find projections of 2k points onto 2k line segments in less than 0.05s.
function [x, y] = projectPointLine(xp, yp, x1, y1, x2, y2)
if nargin > 0
xd = (x2-x1)';
yd = (y2-y1)';
dAB = xd.*xd + yd.*yd;
u = bsxfun(@rdivide, bsxfun(@times, bsxfun(@minus, xp, x1'), xd) + ...
bsxfun(@times, bsxfun(@minus, yp, y1'), yd), dAB);
x = bsxfun(@plus, x1', bsxfun(@times, u, xd));
y = bsxfun(@plus, y1', bsxfun(@times, u, yd));
else
nLine = 3;
nPoint = 2;
xp = rand(nPoint, 1) * 2 -1;
yp = rand(nPoint, 1) * 2 -1;
x1 = rand(nLine, 1) * 2 -1;
y1 = rand(nLine, 1) * 2 -1;
x2 = rand(nLine, 1) * 2 -1;
y2 = rand(nLine, 1) * 2 -1;
tic;
[x, y] = projectPointLine(xp, yp, x1, y1, x2, y2);
toc
close all;
plot([x1'; x2'], [y1'; y2'], '.-', 'linewidth', 2, 'markersize', 20);
axis equal;
hold on
C = lines(nPoint + nLine);
for i=1:nPoint
scatter(x(i, :), y(i, :), 100, C(i+nLine, :), 'x', 'linewidth', 2);
scatter(xp(i), yp(i), 100, C(i+nLine, :), 'x', 'linewidth', 2);
end
for i=1:nLine
scatter(x(:, i)', y(:, i)', 100, C(i, :), 'o', 'linewidth', 2);
end
end
end
回答13:
Just for the sake of completeness, here is a solution using homogeneous coordinates.
The homogeneous points are:
p1 = (x1,y1,1), p2 = (x2,y2,1), p3 = (x3,y3,1)
a line through two points is their cross-product
l_12 := p1 x p2 = (y1-y2, x2-x1, x1*y2 - x2*y1)
The (signed) distance of a point to a line is their dot product.
d := l_12 * p3 = x3*(y1-y2) + y3*(x2-x1) + x1*y2 - x2*y1
The vector from p4 to p3 is d times the normal vector of l_12 divided by the squared length of the normal vector.
n2 := (y1-y2)^2 + (x2-x1)^2
p4 := p3 + d/n2*(y1-y2, x2-x1, 0)
Note: if you divide l_12 by the length of the normal vector
l_12 := l_12 / sqrt((y1-y2)^2 + (x2-x1)^2)
the distance d will be the euclidean distance.
来源：https://stackoverflow.com/questions/1811549/perpendicular-on-a-line-from-a-given-point