Given line equation of 2-order:
$$9 x_{2} + 121 y_{2} - 1089 = 0$$
This equation looks like:
$$a_{11} y_{2}^{2} + 2 a_{12} x_{2} y_{2} + 2 a_{13} y_{2} + a_{22} x_{2}^{2} + 2 a_{23} x_{2} + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{121}{2}$$
$$a_{22} = 0$$
$$a_{23} = \frac{9}{2}$$
$$a_{33} = -1089$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
|a11 a12|
I2 = | |
|a12 a22|
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
|a11 a13| |a22 a23|
K2 = | | + | |
|a13 a33| |a23 a33|
substitute coefficients
$$I_{1} = 0$$
|0 0|
I2 = | |
|0 0|
$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{121}{2}\\0 & 0 & \frac{9}{2}\\\frac{121}{2} & \frac{9}{2} & -1089\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
| 0 121/2| | 0 9/2 |
K2 = | | + | |
|121/2 -1089| |9/2 -1089|
$$I_{1} = 0$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2}$$
$$K_{2} = - \frac{7361}{2}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge \left(I_{1} = 0 \vee K_{2} = 0\right)$$
then by line type:
this equation is of type : two coincident straight lines
$$I_{1} \tilde x2^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
False
None
- reduced to canonical form