**comp.graphics.algorithms**

## Subject: **Re: is a point 'inside' a bezier?**

On 7 May 2006 20:57:03 -0700, daniel.w.gelder@gmail.com wrote:

>Anyhow that equation looks pretty complicated for a simple implicit

>quadratic, especially since the result came out so simple. Can you

>elaborate on what you're doing?

An implicit curve is a concept from algebraic geometry, the set of

points at which a polynomial, or ratio of polynomials, evaluates to

zero. In general degree we have curves that can be represented only

implicitly, not parametrically. For degree 2, we can always choose

either form. Mathematicians prefer to work with projective geometry

and homogeneous coordinates, and so do we in general, not here; but

they also prefer to work with complex numbers, where we prefer reals.

Unlike the complex numbers, the reals are ordered. So we can treat the

implicit function as defining "sides", positive and negative. That's

what you originally asked for.

The only fly in the ointment is that you provide a parametric function

but we need its implicit form. I simply grabbed a conversion formula

from the first source that occurred to me.

Could we write down a formula without the determinants? Yes, but it

turns out to be rather long and messy, and not really that helpful.

For example, we have

L01(x,y) = 2 (-x0 y + x1 y + x y0 - x1 y0 - x y1 + x0 y1)

L12(x,y) = 2 (-x1 y + x2 y + x y1 - x2 y1 - x y2 + x1 y2)

L02(x,y) = -x0 y + x2 y + x y0 - x2 y0 - x y2 + x0 y2

Sorry, I'm not going to explain the theory of implicitization today.

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