Mister Exam

3*x+2 canonical form

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The solution

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2 + 3*x = 0
3x+2=03 x + 2 = 0
3*x + 2 = 0
Detail solution
Given line equation of 2-order:
3x+2=03 x + 2 = 0
This equation looks like:
a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0
where
a11=0a_{11} = 0
a12=0a_{12} = 0
a13=32a_{13} = \frac{3}{2}
a22=0a_{22} = 0
a23=0a_{23} = 0
a33=2a_{33} = 2
To calculate the determinant
Δ=a11a12a12a22\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=0000\Delta = \left|\begin{matrix}0 & 0\\0 & 0\end{matrix}\right|
Δ=0\Delta = 0
Because
Δ\Delta
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
x0=x~cos(ϕ)y~sin(ϕ)x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}
y0=x~sin(ϕ)+y~cos(ϕ)y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}
x0=00x_{0} = 0 \cdot 0
y0=00y_{0} = 0 \cdot 0
x0=0x_{0} = 0
y0=0y_{0} = 0
The center of canonical coordinate system at point O
(0, 0)

Basis of the canonical coordinate system
e1=(1, 0)\vec e_1 = \left( 1, \ 0\right)
e2=(0, 1)\vec e_2 = \left( 0, \ 1\right)
Invariants method
Given line equation of 2-order:
3x+2=03 x + 2 = 0
This equation looks like:
a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0
where
a11=0a_{11} = 0
a12=0a_{12} = 0
a13=32a_{13} = \frac{3}{2}
a22=0a_{22} = 0
a23=0a_{23} = 0
a33=2a_{33} = 2
The invariants of the equation when converting coordinates are determinants:
I1=a11+a22I_{1} = a_{11} + a_{22}
     |a11  a12|
I2 = |        |
     |a12  a22|

I3=a11a12a13a12a22a23a13a23a33I_{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(λ)=a11λa12a12a22λI{\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
I1=0I_{1} = 0
     |0  0|
I2 = |    |
     |0  0|

I3=00320003202I_{3} = \left|\begin{matrix}0 & 0 & \frac{3}{2}\\0 & 0 & 0\\\frac{3}{2} & 0 & 2\end{matrix}\right|
I(λ)=λ00λI{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda\end{matrix}\right|
     | 0   3/2|   |0  0|
K2 = |        | + |    |
     |3/2   2 |   |0  2|

I1=0I_{1} = 0
I2=0I_{2} = 0
I3=0I_{3} = 0
I(λ)=λ2I{\left(\lambda \right)} = \lambda^{2}
K2=94K_{2} = - \frac{9}{4}
Because
I2=0I3=0(I1=0K2=0)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
I1y~2+K2I1=0I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0
or
False

None

- reduced to canonical form