Perpendicular on a line from a given point

Question

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).

Answer 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

Answer 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)

Answer 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;

Answer 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);
}

Answer 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)

Answer 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

Answer 7

Find out the slopes for both the lines, say slopes are m1 and m2 then m1*m2=-1 is the condition for perpendicularity.

Answer 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

Answer 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}]

Answer 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}]

Answer 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

Answer 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