How do I calculate the normal vector of a line segment?

Question

Suppose I have a line segment going from (x1,y1) to (x2,y2). How do I calculate the normal vector perpendicular to the line?

I can find lots of stuff about doing this for planes in 3D, but no 2D stuff.

Please go easy on the maths (links to worked examples, diagrams or algorithms are welcome), I'm a programmer more than I'm a mathematician ;)

Answer 1

If we define dx = x2 - x1 and dy = y2 - y1, then the normals are (-dy, dx) and (dy, -dx).

Note that no division is required, and so you're not risking dividing by zero.

Answer 2

Another way to think of it is to calculate the unit vector for a given direction and then apply a 90 degree counterclockwise rotation to get the normal vector.

The matrix representation of the general 2D transformation looks like this:

x' = x cos(t) - y sin(t)
y' = x sin(t) + y cos(t)

where (x,y) are the components of the original vector and (x', y') are the transformed components.

If t = 90 degrees, then cos(90) = 0 and sin(90) = 1. Substituting and multiplying it out gives:

x' = -y
y' = +x

Same result as given earlier, but with a little more explanation as to where it comes from.

Answer 3

This question has been posted long time ago, but I found an alternative way to answer it. So I decided to share it here.
Firstly, one must know that: if two vectors are perpendicular, their dot product equals zero.
The normal vector (x',y') is perpendicular to the line connecting (x1,y1) and (x2,y2). This line has direction (x2-x1,y2-y1), or (dx,dy).
So,

(x',y').(dx,dy) = 0
x'.dx + y'.dy = 0

The are plenty of pairs (x',y') that satisfy the above equation. But the best pair that ALWAYS satisfies is either (dy,-dx) or (-dy,dx)

Answer 4

m1 = (y2 - y1) / (x2 - x1)

if perpendicular two lines:

m1*m2 = -1

then

m2 = -1 / m1 //if (m1 == 0, then your line should have an equation like x = b)

y = m2*x + b //b is offset of new perpendicular line.. 

b is something if you want to pass it from a point you defined