Find point perpendicular to line

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Jeff Higgins - 22 Jul 2007 18:23 GMT

Hi,
My trigonometry and imagination are failing me now.
I hope some one can give me help.

I have two points:

Point2D.Double A = new Point2D.Double(100d, 100d);
Point2D.Double B = new Point2D.Double(200d, 200d);

which can be used to describe a line.

Path2D.Double lineAB = new Path2D.Double();
moveTo(A.x, A.y);
lineTo(B.x, B.y);

I would like to find a third point that is:

on a line perpendicular to lineAB,
and
double distanceC = 3d;
distant from A

double distanceB = Math.sqrt(Math.pow((B.x - A.x), 2) + Math.pow((B.y -
A.y), 2));

My ultimate goal is to build a rectangle:
Rectangle2D.Double hitTestBounds;
that encloses lineAB such that lineAB splits
hitTestBounds in half, not diagonally.

Thanks,
Jeff Higgins

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Lew - 22 Jul 2007 18:40 GMT

> I would like to find a third point that is:
>
> on a line perpendicular to lineAB,
> and
> double distanceC = 3d;
> distant from A

Every point on the circumference of a circle of radius 3d centered at A
fulfills your requirements.

If you further constrain the point to be at distance 3d from both points A and
B, then you have exactly two points that fulfill the requirements, those at
the intersections of the two circles of radius 3d centered at A and B.

It is meaningless to describe a point as "perpendicular to a line", as your
subject line requests. Every point in the plane is on a line perpendicular to
AB, as requested in your message body.

Signature

Lew

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Jeff Higgins - 22 Jul 2007 19:13 GMT

>> I would like to find a third point that is:
>>
>> on a line perpendicular to lineAB,

and passes throught point A

>> and
>> double distanceC = 3d;
[quoted text clipped - 11 lines]
> your subject line requests. Every point in the plane is on a line
> perpendicular to AB, as requested in your message body.

Reply to this Message

Joshua Cranmer - 22 Jul 2007 19:45 GMT

>>> I would like to find a third point that is:
>>>
[quoted text clipped - 5 lines]
>>> double distanceC = 3d;
>>> distant from A

I belive this is your criteria:
C
| -- distance 3d.
B-----A

If the line AB is at an angle theta to the origin line, then theta =
arctan (Ay-By)/(Ax-Bx). The angle that line AC makes with the origin is
phi, which is theta + pi/2 in this case. Translating the origin to point
A, we have that Cx is r*cos phi and Cy is r*sin phi.

phi should also be equal to arctan (Bx-Ax)/(Ay-By), and reducing, we get
the Cx = 3d / sqrt (1+x^2) + Ax and Cy = 3d * x /sqrt(1+x^2) + Ay where x
is (Bx-Ax)/(Ay-By), except if Ay=By (i.e., a horizontal line), where Cx =
Ax and Cy = Ay+3d.

Note: only checked for slope AB = 0, 1, and infinity.

Reply to this Message

Jeff Higgins - 23 Jul 2007 01:41 GMT

>>>> I would like to find a third point that is:
>>>>
[quoted text clipped - 22 lines]
>
> Note: only checked for slope AB = 0, 1, and infinity.

Joshua,
Thanks very much.
Appreciative,
JH

import java.awt.*;
import java.awt.geom.*;
import javax.swing.*;

public class HitTestBoundsTest
{
static class GraphicPanel extends JPanel
{
Point2D.Double getMyPoint_PerpendicularToALine(
Point2D.Double source,
Point2D.Double target,
double width)
{
// TODO test for horizontal line
double x = ((target.x - source.x) / (source.y - target.y));
double cx = (width / 2) / Math.sqrt(1 + Math.pow(x, 2)) + source.x;
double cy = (width / 2) * x / Math.sqrt(1 + Math.pow(x, 2)) +
source.y;
Point2D.Double ret = new Point2D.Double(cx, cy);
return ret;
}

public void paintComponent( Graphics g )
{
Graphics2D g2 = (Graphics2D)g;
Path2D.Double line = new Path2D.Double();
line.moveTo(source.x, source.y);
line.lineTo(target.x, target.y);
Point2D.Double C =
getMyPoint_PerpendicularToALine(source, target, width);
Point2D.Double D =
getMyPoint_PerpendicularToALine(target, source, width);
Path2D.Double CD = new Path2D.Double();
Path2D.Double EF = new Path2D.Double();
Path2D.Double CE = new Path2D.Double();
Path2D.Double DF = new Path2D.Double();
CD.moveTo(C.x, C.y);
CD.lineTo(D.x, D.y);
EF.moveTo(C.y, C.x);
EF.lineTo(D.y, D.x);
CE.moveTo(C.x, C.y);
CE.lineTo(C.y, C.x);
DF.moveTo(D.x, D.y);
DF.lineTo(D.y, D.x);
g2.draw(CD);
g2.draw(EF);
g2.draw(CE);
g2.draw(DF);
g2.setColor(Color.RED);
g2.draw(line);
}

public GraphicPanel(){}

Point2D.Double source =
new Point2D.Double(100d, 100d);
Point2D.Double target =
new Point2D.Double(200d, 200d);
double width = 3d;
}

JFrame frame;
JPanel panel;

HitTestBoundsTest()
{
frame = new JFrame("HitTestBoundsTest");
panel = new GraphicPanel();
frame.setSize(600, 480);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.add(panel);
frame.setVisible(true);
}

public static void main(String[] args)
{
HitTestBoundsTest test = new HitTestBoundsTest();
}
}

Reply to this Message

Mark Space - 22 Jul 2007 20:47 GMT

> It is meaningless to describe a point as "perpendicular to a line", as
> your subject line requests. Every point in the plane is on a line
> perpendicular to AB, as requested in your message body.

Unless, I think, the three points are co-linear, including the case
where one or more points are the same.

Reply to this Message

Lew - 22 Jul 2007 21:56 GMT

Lew wrote:

>> It is meaningless to describe a point as "perpendicular to a line", as
>> your subject line requests. Every point in the plane is on a line
>> perpendicular to AB, as requested in your message body.

> Unless, I think, the three points are co-linear, including the case
> where one or more points are the same.

A and B must be distinct points, of course, else they would not define a line.
Every point on the plane, including the points A and B themselves and any
point on AB, is therefore located on some perpendicular to AB.

Signature

Lew

Reply to this Message

Mark Space - 22 Jul 2007 20:48 GMT

> My ultimate goal is to build a rectangle:
> Rectangle2D.Double hitTestBounds;
> that encloses lineAB such that lineAB splits
> hitTestBounds in half, not diagonally.

Just curious: you say line, but do you mean perhaps line segment? Is
this line AB infinite, or does it have a finite length, starting at A
and ending a B?

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Find point perpendicular to line

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