Given line equation of 2-order:
$$x^{2} + 5 x = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = \frac{5}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(\tilde x + \frac{5}{2}\right)^{2} = \frac{25}{4}$$
$$\tilde x'^{2} = \frac{25}{4}$$
Given equation is two parallel straight lines
- reduced to canonical form
where replacement made
$$\tilde x' = \tilde x + \frac{5}{2}$$
$$\tilde y' = \tilde y$$
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = \left(- \frac{5}{2}\right) 1 + 0 \cdot 0$$
$$y_{0} = \left(- \frac{5}{2}\right) 0 + 0 \cdot 1$$
$$x_{0} = - \frac{5}{2}$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
(-5/2, 0)
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$