How can you determine a point is between two other

Question

Let's say you have a two dimensional plane with 2 points (called a and b) on it represented by an x integer and a y integer for each point.

How can you determine if another point c is on the line segment defined by a and b?

I use python most, but examples in any language would be helpful.

Answer 1

Using a more geometric approach, calculate the following distances:

ab = sqrt((a.x-b.x)**2 + (a.y-b.y)**2)
ac = sqrt((a.x-c.x)**2 + (a.y-c.y)**2)
bc = sqrt((b.x-c.x)**2 + (b.y-c.y)**2)

and test whether ac+bc equals ab:

is_on_segment = abs(ac + bc - ab) < EPSILON

That's because there are three possibilities:

• The 3 points form a triangle => ac+bc > ab
• They are collinear and c is outside the ab segment => ac+bc > ab
• They are collinear and c is inside the ab segment => ac+bc = ab

Answer 2

I needed this for javascript for use in an html5 canvas for detecting if the users cursor was over or near a certain line. So I modified the answer given by Darius Bacon into coffeescript:

is_on = (a,b,c) ->
    # "Return true if point c intersects the line segment from a to b."
    # (or the degenerate case that all 3 points are coincident)
    return (collinear(a,b,c) and withincheck(a,b,c))

withincheck = (a,b,c) ->
    if a[0] != b[0]
        within(a[0],c[0],b[0]) 
    else 
        within(a[1],c[1],b[1])

collinear = (a,b,c) ->
    # "Return true if a, b, and c all lie on the same line."
    ((b[0]-a[0])*(c[1]-a[1]) < (c[0]-a[0])*(b[1]-a[1]) + 1000) and ((b[0]-a[0])*(c[1]-a[1]) > (c[0]-a[0])*(b[1]-a[1]) - 1000)

within = (p,q,r) ->
    # "Return true if q is between p and r (inclusive)."
    p <= q <= r or r <= q <= p

Answer 3

Here's another approach:

• Lets assume the two points be A (x1,y1) and B (x2,y2)
• The equation of the line passing through those points is (x-x1)/(y-y1)=(x2-x1)/(y2-y1) .. (just making equating the slopes)

Point C (x3,y3) will lie between A & B if:

• x3,y3 satisfies the above equation.
• x3 lies between x1 & x2 and y3 lies between y1 & y2 (trivial check)

Answer 4

The scalar product between (c-a) and (b-a) must be equal to the product of their lengths (this means that the vectors (c-a) and (b-a) are aligned and with the same direction). Moreover, the length of (c-a) must be less than or equal to that of (b-a). Pseudocode:

# epsilon = small constant

def isBetween(a, b, c):
    lengthca2  = (c.x - a.x)*(c.x - a.x) + (c.y - a.y)*(c.y - a.y)
    lengthba2  = (b.x - a.x)*(b.x - a.x) + (b.y - a.y)*(b.y - a.y)
    if lengthca2 > lengthba2: return False
    dotproduct = (c.x - a.x)*(b.x - a.x) + (c.y - a.y)*(b.y - a.y)
    if dotproduct < 0.0: return False
    if abs(dotproduct*dotproduct - lengthca2*lengthba2) > epsilon: return False 
    return True

Answer 5

You can use the wedge and dot product:

def dot(v,w): return v.x*w.x + v.y*w.y
def wedge(v,w): return v.x*w.y - v.y*w.x

def is_between(a,b,c):
   v = a - b
   w = b - c
   return wedge(v,w) == 0 and dot(v,w) > 0

Answer 6

Check if the cross product of (b-a) and (c-a) is 0, as tells Darius Bacon, tells you if the points a, b and c are aligned.

But, as you want to know if c is between a and b, you also have to check that the dot product of (b-a) and (c-a) is positive and is less than the square of the distance between a and b.

In non-optimized pseudocode:

def isBetween(a, b, c):
    crossproduct = (c.y - a.y) * (b.x - a.x) - (c.x - a.x) * (b.y - a.y)

    # compare versus epsilon for floating point values, or != 0 if using integers
    if abs(crossproduct) > epsilon:
        return False

    dotproduct = (c.x - a.x) * (b.x - a.x) + (c.y - a.y)*(b.y - a.y)
    if dotproduct < 0:
        return False

    squaredlengthba = (b.x - a.x)*(b.x - a.x) + (b.y - a.y)*(b.y - a.y)
    if dotproduct > squaredlengthba:
        return False

    return True