## 2 Materials and Methods
### The Exact (Mathematical) Method for SSTPC2L  
For this method, in order to simplify calculus, we apply some initial transformations (rotations and translations) to the points A\({}_{11}\), A\({}_{12}\), A\({}_{21}\), A\({}_{22}\), and Q so that the bisector of the angle between the lines d\({}_{1}\) and d\({}_{2}\) becomes parallel to Oy axis and the point Q is in the origin O, i.e., Q = O (0,0). To do so, we first calculate the intersection I of d\({}_{1}\) and d\({}_{2}\):  
\[\text{d}_{1}\cap\text{ d}_{2}\ =\ \{\text{I}\} \tag{7}\]  
by solving the system of linear equations:  
\[\left\{ \begin{array}{c}\text{d}_{1}:\frac{x-x_{11}}{x_{12}-x_{11}}\ =\ \frac{y-y_{11}}{y_{12}-y_{11}}\\ \text{d}_{2}:\frac{x-x_{21}}{x_{22}-x_{21}}\ =\ \frac{y-y_{21}}{y_{22}-y_{21}} \end{array} \right.\]  
The following transformations to A\({}_{11}\), A\({}_{12}\), A\({}_{21}\), A\({}_{22}\) and Q are applied:  
1. Translation with \((-x_{\text{I}}\), \(-y_{\text{I}})\) (I is moved into origin O)
2. Rotation with \(\sin(-u)=-\sin(u)\) and \(\cos(-u)=\cos(u)\), where: \[\left\{ \begin{array}{c}\text{r}\ =\ \sqrt{\left(x_{12}-x_{11}\right)^{2}+ \left(y_{12}-y_{11}\right)^{2}}\\ \cos(u)\ =\ \left(x_{12}-x_{11}\right)/\text{r}\\ \sin(u)\ =\ \left(y_{12}-y_{11}\right)/\text{r}\end{array} \right.\] (8) After this rotation, the line d\({}_{1}\) is over Ox axis of the map's system of coordinates. 3. Rotation with \(\sin(v)\) and \(\cos(v)\), where: \[\left\{ \begin{array}{c}\text{w}\ =\ \text{atan2}(y_{22}-y_{21},\ x_{22}-x_{21})\\ \cos(v)\ =\ \cos(\pi/2-w/2)\\ \sin(v)\ =\ \sin(\pi/2-w/2)\end{array} \right.\] (9) After this rotation, the bisector of the angle given by lines d\({}_{1}\) and d\({}_{2}\) is over Oy. 4. Translation with \((-x_{\text{Q}}\), \(-y_{\text{Q}})\) (Q is moved into origin O)  
The following 2 systems of equations have to be solved in order to find the points P\({}_{\text{i}}\)(x\({}_{\text{i}}\),y\({}_{\text{i}}\)) (i = 1,2) where the lines d\({}_{\text{i}}\) (i = 1,2) intersect the line d:  
\[\left\{ \begin{array}{c}\text{y}\ =\ \text{a}_{\text{i}}\text{x}+\text{b}_{ \text{i}}\\ \text{y}\ =\ \text{ax}\end{array} \right. \tag{10}\]  
Since A\({}_{\text{i}1}\), A\({}_{\text{i}2}\) define the line d\({}_{\text{i}}\) (i = 1,2) in (10) we have:  
\[\text{a}_{\text{i}}\ =\ \frac{(y_{\text{i}2}-y_{\text{i}1})}{(x_{\text{i}2}-x_{ \text{i}1})} \tag{11}\]  
\[\text{b}_{\text{i}}\ =\ y_{\text{i}1}-\text{a}_{\text{i}1}x_{\text{i}1} \tag{12}\]  
Solving the 2 systems of equations, the points P\({}_{\text{i}}\)(x\({}_{\text{i}}\),y\({}_{\text{i}}\)) (i = 1,2) are found, where:  
\[\text{x}_{\text{i}}\ =\ \frac{\text{b}_{\text{i}}}{\text{a}-\text{a}_{\text{i}}} \tag{13}\]\[y_{1}\ =\ \frac{ab_{i}}{a-a_{i}} \tag{14}\]  
The distance between \(P_{1}\) and \(P_{2}\) must be minimized, i.e.,:  
\[\min_{a}f(a) \tag{15}\]  
where:  
\[f(a)\ =\ \left(y_{2}-y_{1}\right)^{2}+\left(x_{2}-x_{1}\right)^{2} \tag{16}\]  
By replacing (13) and (14) in (16) the following formula for f(a) is obtained:  
\[f(a)\ =\ \left(\frac{a_{2}b_{2}}{a-a_{2}}-\frac{a_{1}b_{1}}{a-a_{1}}+b_{2}-b_{1} \right)^{2}+\left(\frac{b_{2}}{a-a_{2}}-\frac{b_{1}}{a-a_{1}}\right)^{2} \tag{17}\]  
We make the following notations:  
\[a_{L}\ =\ \min\left\{\frac{-1}{a_{1}},\frac{-1}{a_{2}}\right\} \tag{18}\]  
\[a_{R}\ =\ \max\left\{\frac{-1}{a_{1}},\frac{-1}{a_{2}}\right\} \tag{19}\]  
It is easy to see that the function \(f\) is decreasing for \(a\leq a_{L}\), and is increasing for \(a\geq a_{R}\), since \(f\) is continuous, \(a_{L}\) and \(a_{R}\) are the slopes of the perpendiculars \(QQ_{1}\) and \(QQ_{2}\) from \(Q\) on the lines \(d_{1}\) and, respectively, \(d_{2}\), and \"\(a\)\" is the slope of the line \(d_{3}\). Consequently, there is a minimum of the function \(f\) denoted \(a_{min}\) in the interval \([a_{L},a_{R}]\). So, the value \(a_{min}\) is one of the solutions of the equation:  
\[f^{\prime}(a)\ =\ 0 \tag{20}\]  
on the interval \([a_{L},a_{R}]\), where:  
\[\text{b}_{\text{i}}\ =\ y_{\text{i}1}-\text{a}_{\text{i}1}x_{\text{i}1} \tag{12}\]0  
**Remark 1.