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How to use it?
- Canonical form:
- y^2-8x+2y+17=0
- x^2+y^2+4z^2+2xy+4xz-4yz-2x+2y+2z=0
- 2x^2+6z^2+y^2-4x-4z+8=0
- 2x^2-8xy+8y^2-4x+8y-3=0
- Plot:
- z=sqrt(x*y)/x^2+y^2
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3x^2+2y^2+4xy-x
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3*cos(x)*sin(y)-sin(y)*cos(y)-4*cos(y)-3*sqrt(3)=0
- 2*(3*x+4)+12=5*y
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Identical expressions
- (three x+ two)*(x-3)
- (3x plus 2) multiply by (x minus 3)
- (three x plus two) multiply by (x minus 3)
- (3x+2)(x-3)
- 3x+2x-3
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Similar expressions
- (3x+2)*(x+3)
- (3x-2)*(x-3)
(3x+2)*(x-3) canonical form
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The solution
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(-3 + x)*(2 + 3*x) = 0
$$\left(x - 3\right) \left(3 x + 2\right) = 0$$
(x - 3)*(3*x + 2) = 0
Detail solution
Given line equation of 2-order:
$$\left(x - 3\right) \left(3 x + 2\right) = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = - \frac{7}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -6$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$\left(x - 3\right) \left(3 x + 2\right) = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = - \frac{7}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -6$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
Invariants method
Given line equation of 2-order:
$$\left(x - 3\right) \left(3 x + 2\right) = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = - \frac{7}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -6$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 3$$
$$I_{3} = \left|\begin{matrix}3 & 0 & - \frac{7}{2}\\0 & 0 & 0\\- \frac{7}{2} & 0 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 3$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 3 \lambda$$
$$K_{2} = - \frac{121}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge K_{2} < 0 \wedge I_{1} \neq 0$$
then by line type:
this equation is of type : two parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$3 \tilde y^{2} - \frac{121}{12} = 0$$
- reduced to canonical form
$$\left(x - 3\right) \left(3 x + 2\right) = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = - \frac{7}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -6$$
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
$$I_{1} = 3$$
|3 0|
I2 = | |
|0 0|$$I_{3} = \left|\begin{matrix}3 & 0 & - \frac{7}{2}\\0 & 0 & 0\\- \frac{7}{2} & 0 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
| 3 -7/2| |0 0 |
K2 = | | + | |
|-7/2 -6 | |0 -6|$$I_{1} = 3$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 3 \lambda$$
$$K_{2} = - \frac{121}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge K_{2} < 0 \wedge I_{1} \neq 0$$
then by line type:
this equation is of type : two parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$3 \tilde y^{2} - \frac{121}{12} = 0$$
None
- reduced to canonical form