1st quadrant proofs in coordinate geometry

Question

A similar question has already been asked here.

After going through the answers, I wanted to know:

If I were to prove something in coordinate geometry, (for eg: a general formula,etc) will proving it in the first quadrant give us a general result? Will it apply to all cases?

For eg: Say I've to prove that the area of a triangle is given by: $$ Area({\triangle})=\frac{1}{2}|(x_{1}y_{2} + x_2y_3 + x_3y_1)-(x_2y_1 + x_3y_2 + x_1y_3)| $$

If I prove this only in the 1st quadrant, will the result be a general result?

If the answer for the above question is Yes, then is there any "proof" to prove that anything we prove in the first quadrant applies to all quadrants, i.e, it is a general result?

Answer 1

Note that the square of area$(\Delta)$ is a polynomial in the coordinates $x_i$, $y_i$. It can be seen as a function of the three points, so as a function defined on ${\bf R}^2\times {\bf R}^2\times{\bf R}^2$.

If you have two polynomials that are equal in the first quadrant, they are equal everywhere. This comes from the fact that the set of points where a non-constant polynomial vanishes is of lower dimension than the space on which it is defined. Hence it cannot contain an non-empty open set without being equal to the whole set.

In your example, it is known a priori that the square of the area is a polynomial because it is given by a determinant (a Gram determinant to be precise). So showing the formula mentioned in your question for a triple of vectors all belonging to the quarter plan is enough to ensure that it is valid everywhere.

There are many quantities in plane geometry that are related to polynomials, but not all. As shown in the comments, one must be cautious when dealing with the absolute values, signs, inequalities and orientation. For example, the non-oriented angle between a vector in the quarter plane and the x-axis is between $0$ and $\pi \over 2$ but this does not extend to all vectors of the plane. Here the angle is given by $\Bigl| \arccos\Bigl({x \over \sqrt{x^2+y^2}}\Bigr) \Bigr|$. This is definitely not a polynomial.