CODEWITHSUNDEEP
c++geometry
feelfree
feelfree

Reputation: 11753

What's the best way to know the relative position of a point with regard to a straight line?

Suppose now I have a straight line in the 2D coordinate system, which can be wrote as:

ax+by+c=0

Now I have a point, whose coordinate is (x0,y0), and my question is how I can tell the relative position of the point with regard to the line (above or below). One solution I can figure out is as follows:

// set up the line
a = ..;
b = ..;
c = ..;
// make sure that b is negative
if (b>0)
{
  a = -a;
  b = -b;
  c= -c;
}
// set the point
x0 = ..;
y0 = ..;

temp = a*x0+b*y0+c;

if (temp<0)
// the point is above the line
else
// the point is under the line.

In the above codes I did not consider the special case of straight line (horizontal or vertical), but the basic idea is to make sure b is negative first, and then calculate the function's value (temp), and depending on its value determine the point's position. I am not sure whether this is a good solution. Any ideas will be appreciated.

Upvotes: 2

Views: 2423

Answers (2)

MBo
MBo

Reputation: 80197

Yes, your solution is right. Possible optimization - don't negate coefficients, just evaluate sign of expression b*(a*x+b*y+c). + for above, - for under, 0 for points on line and for indefinite case - vertical lines.

Explanation: common line equation

a*x+b*y+c=0

may be transformed to normal equation

x*cos(t)+y*sin(t)-p=0

(with division by -sign(c)*Sqrt(a^2+b^2))
where p is the length of normal to line from coordinate origin (O), t is angle between OX axis and this normal.

The sign of expression 

XX*cos(t)+YY*sin(t)-p=0

is positive, when point (XX, YY) and coordinate origin are divided by this line, and negative otherwise. So for t in range 0..Pi sin is positive, and positive value of expression denotes that point is above the line (origin is under) and so on...

Upvotes: 3

laune
laune

Reputation: 31290

Corrected version

You don't have to invert any signs. Simply inserting the coordinates of the point (x,y) into the form of the line ax + by + c results in a value, call this h. Also, compute d = -c/b. If h and d have the same sign and if 'd > 0', then (x,y) is "below".

bool isBelow( double a, double b, double c, double x0, double y0 ){
  double d = -c/b;
  if( d == 0 ){
    return -a/b*x0 > y0;
  } else {
    double h0 = a*x0 + b*y0 + c;
    return d > 0 && (d > 0 == h0 > 0);
  }
}

It will also work for for horizontal lines, but not for vertical lines (where b==0).

Upvotes: 2

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