5*x+1 canonical form

The teacher will be very surprised to see your correct solution 😉

Detail solution

Given line equation of 2-order:

5 x + 1 = 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} = 0

a_{12} = 0

a_{13} = \frac{5}{2}

a_{22} = 0

a_{23} = 0

a_{33} = 1

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}0 & 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)