Is there a single equation describing a triangle in the Cartesian plane, using only arithmetic operations? (Yes.)

Question

Is there a single equation describing a triangle in the Cartesian plane which includes only arithmetic operations?

The below uses the square root---but, in this case, the square root is itself may be computed using only arithmetic.

This post is in response to Is there an equation for a triangle? by "Stand-up Maths" (2021-12-01). Therein, a single equation for a triangle in the Cartesian plane is presented (at about minute 13). However, that equation is not valid if any of the sides of the triangle are parallel to the y-axis. I present the following as an alternative. See also "Is there any equation for triangle?" where the questioner starts off with the concept of slope, which is undefined for "vertical" lines.

Let there be three mutually non-parallel lines $l_1$, $l_2$, $l_3$ in the Cartesian Plane:

$$c_1 + a_1 x + b_1 y = 0$$ $$c_2 + a_2 x + b_2 y = 0$$ $$c_3 + a_3 x + b_3 y = 0$$

respectively, $c_k, a_k, b_k \in \mathbf{Z}$ (i.e., the integers); $k \in \left\{ 1, 2, 3\right\}$

Let $P$ be an arbitrary point $(x, y); x, y \in \mathbf{Q}$ (i.e., the rational numbers).

Let $Q_1$, $Q_2$, and $Q_3$ be the quadrances ("square distances") from $P$ to lines $l_1$, $l_2$, and $l_3$, respectively.

$P$ is on at least one of the three lines if and only if

$$Q_1 Q_2 Q_3 = 0$$

Recall that $0 \le Q_1 Q_2 Q_3$.

Let $z_1$, $z_2$, and $z_3$ be points on two lines (a.k.a., the intersection of the two lines): $$z_1 = l_2 l_3,$$ $$z_2 = l_3 l_1,$$ $$z_3 = l_1 l_2.$$

Let triangles $T_1$, $T_2$, and $T_3$ be the triangles formed by: $P, z_2, z_3$; $P, z_1, z_2$; and $P, z_3, z_1$; respectively. Let triangle $T$ be the triangle formed by $z_1$, $z_2$, and $z_3$ (i.e., by $l_1$, $l_2$, and $l_3$).

Let $A_1$, $A_2$, $A_3$ be the quadreas ("square areas," basically) of triangles $T_1$, $T_2$, $T_3$.

Let $A$ be the quadrea of triangle $T$.

Then $P$ is either on triangle $T$ or in the triangular region bounded by $T$ if and only if

$$\sqrt{A_1} +\sqrt{A_2} +\sqrt{A_3} -\sqrt{A} = 0$$

(Note that the expression on the left hand side cannot be negative. Note that each of the terms is rational.)

Then an equation of a triangle in the Cartesian Plane is:

$$Q_1 Q_2 Q_3 + \sqrt{A_1} +\sqrt{A_2} +\sqrt{A_3} -\sqrt{A} = 0$$

or

$$Q_1 Q_2 Q_3 = \sqrt{A} -\sqrt{A_1} -\sqrt{A_2} -\sqrt{A_3}$$

Either of the above equations could be expanded into an equation in $c_k$, $a_k$, $b_k$, $x$, and $y$ only and, with the exception of the square root, would include only arithmetic operations. Moreover, I claim that, under the constraints laid out above, the "$\sqrt{}$" terms are arithmetic, since there exists at least one arithmetic process for determining exactly the square root of any rational number.

Addendum I: The "Is there an equation for a triangle?" equation for a triangle

In response to @Blue 's 2021-12-13 comment, "You should include the solution presented by Stand-up Maths," that solution is:

$$(\mathrm{sign}(m_1x-y)+1)(\mathrm{sign}(m_2x-y)-1)(a_1x+b_1-y)\\+(\mathrm{sign}(m_2x-y)+1)(\mathrm{sign}(m_3x-y)+1)(a_2x+b_2-y)\\+(\mathrm{sign}(m_3x-y)-1)(\mathrm{sign}(m_1x-y)-1)(a_3x+b_3-y)=0$$

(transcribed from "Is there an equation for a triangle?" [about minute 13])).

Answer 1

You can't do it using only $+-\times\div$. If you make any function using only those, you can write it in the form $\frac{p(x_1,y_1,x_2,y_2,x_3,y_3)}{q(x_1,y_1,x_2,y_2,x_3,y_3)}$ for some polynomials $p(x_1,y_1,x_2,y_2,x_3,y_3)$ and $q(x_1,y_1,x_2,y_2,x_3,y_3)$. Therefore, it is zero only where $p(x_1,y_1,x_2,y_2,x_3,y_3)=0$. But if a polynomial is equal to zero on a 6-dimensional hypercube in $\mathbb{R}^6$, then it has to be zero everywhere in $\mathbb{R}^6$.