Theorem: Characteristic of Quadratic Equation that Represents Two Straight Lines

Consider the '''quadratic equation in $2$ variables''':
:$(1): \quad a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
where $x$ and $y$ are independent variables.
Then $(1)$ represents $2$ straight lines {{iff}} its discriminant equals zero:
:$a b c + 2 f g h - a f^2 - b g^2 - c h^2 = 0$
This can also be expressed in the form of a determinant:
:$\begin {vmatrix} a & h & g \\ h & b & f \\ g & f & c \end {vmatrix} = 0$

Proof:
Suppose that $a \ne 0$.
We have:
{{begin-eqn}}
{{eqn | l = a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^2 x^2 + a b y^2 + 2 a h x y + 2 a g x + 2 a f y + c
      | r = 0
      | c = multiplying by $a$
}}
{{eqn | ll= \leadsto
      | l = \paren {a x + h y + g}^2 + a b y^2 + 2 a f y + a c
      | r = 2 g h y + h^2 y^2 + g^2
      | c = completing the square in $x$ terms
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = \paren {a x + h y + g}^2 - \paren {\paren {h^2 - a b} y^2 + 2 \paren {g h - a f} y + \paren {g^2 - a c} }
      | r = 0
      | c = rearranging
}}
{{end-eqn}}
In order that the second part is a perfect square in $y$, it is necessary that:
{{begin-eqn}}
{{eqn | l = \paren {g h - a f}^2
      | r = \paren {h^2 - a b} \paren {g^2 - a c}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = g^2 h^2 - 2 a f g h + a^2 f^2
      | r = g^2 h^2 - a b g^2 - a c h^2 + a^2 b c
      | c = multiplying out
}}
{{eqn | n = 3
      | ll= \leadsto
      | l = a b c + 2 f g h - a f^2 - b g^2 - c h^2
      | r = 0
      | c = simplifying, rearranging and dividing by $a$ which is non-zero
}}
{{end-eqn}}
Conversely, if $(3)$ is true, then $(2)$ can be expressed in the form of a Difference of Two Squares:
{{begin-eqn}}
{{eqn | n = 2
      | l = \paren {a x + h y + g}^2 - \paren {\paren {h^2 - a b} y^2 + 2 \paren {g h - a f} y + \paren {g^2 - a c} }
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {a x + h y + g}^2 - \paren {h^2 - a b} \paren {y^2 + 2 \dfrac {g h - a f} {h^2 - a b} y + \dfrac {g^2 - a c} {h^2 - a b} }
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {a x + h y + g}^2 - \paren {h^2 - a b} \paren {\paren {y + \dfrac {g h - a f} {h^2 - a b} }^2 + \dfrac {g^2 - a c} {h^2 - a b} - \paren {\dfrac {g h - a f} {h^2 - a b} }^2}
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {a x + h y + g}^2 - \paren {h^2 - a b} \paren {y + \dfrac {g h - a f} {h^2 - a b} }^2
      | r = 0
      | c = as $\paren {g h - a f}^2 - \paren {h^2 - a b} \paren {g^2 - a c}$
}}
{{end-eqn}}
Hence $(2)$ has $2$ factors, which can be seen to be the equations of straight lines.
{{qed|lemma}}
Let $a = 0$ but $b \ne 0$.
Then: 
{{begin-eqn}}
{{eqn | l = b y^2 + 2 h x y + 2 g x + 2 f y + c
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b^2 y^2 + 2 b h x y + 2 b g x + 2 b f y + b c
      | r = 0
      | c = multiplying by $b$
}}
{{eqn | ll= \leadsto
      | l = \paren {b y + h x + f}^2 + 2 b g x + b c
      | r = 2 f h x + h^2 x^2 + f^2
      | c = completing the square in $y$ terms
}}
{{eqn | ll= \leadsto
      | l = \paren {b y + h x + f}^2 - \paren {h^2 x^2 + 2 \paren {f h - b g} x + \paren {f^2 - b c} }
      | r = 0
      | c = rearranging
}}
{{end-eqn}}
In order that the second part is a perfect square in $x$, it is necessary that:
{{begin-eqn}}
{{eqn | l = \paren {f h - b g}^2
      | r = h^2 \paren {f^2 - b c}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = f^2 h^2 - 2 b f g h + b^2 g^2
      | r = f^2 h^2 - b c h^2
      | c = multiplying out
}}
{{eqn | n = 4
      | ll= \leadsto
      | l = 2 f g h - b g^2 - c h^2
      | r = 0
      | c = simplifying, rearranging and dividing by $b$ which is non-zero
}}
{{end-eqn}}
it is noted that $(4)$ is the same as $(3)$ but with $a = 0$.
{{qed|lemma}}
Suppose $a = 0$ and $b = 0$ but $h \ne 0$.
Then:
{{begin-eqn}}
{{eqn | l = 2 h x y + 2 g x + 2 f y + c
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 h^2 x y + 2 g h x + 2 f h y + c h
      | r = 0
      | c = multiplying by $h$
}}
{{eqn | n = 5
      | ll= \leadsto
      | l = 2 \paren {h x + f} \paren {h y + g} + c h
      | r = 2 f g
      | c = extracting factors and completing rectangle
}}
{{end-eqn}}
and it is seen that in order for $(1)$ to be divisible into the $2$ required factors:
:$2 \paren {h x + f} \paren {h y + g} = 0$
it is necessary for $c h = 2 f g$.
This is again the same as $(3)$ when you set $a = 0$ and $b = 0$.
{{qed|lemma}}
If $a = 0$ and $b = 0$ and $h = 0$, then $(1)$ is not a quadratic equation.
All cases have been covered.
{{qed|lemma}}
Finally we see that:
{{begin-eqn}}
{{eqn | l = \begin {vmatrix} a & h & g \\ h & b & f \\ g & f & c \end {vmatrix}
      | r = a \begin {vmatrix} b & f \\ f & c \end {vmatrix} - h \begin {vmatrix} h & f \\ g & c \end {vmatrix} + g \begin {vmatrix} h & b \\ g & f \end {vmatrix}
      | c = Determinant of Order 3
}}
{{eqn | r = a \paren {b c - f^2} - h \paren {h c - f g} + g \paren {h f - b g}
      | c = Determinant of Order 2
}}
{{eqn | r = a b c + 2 f g h - a f^2 - b g^2 - c h^2
      | c = simplifying
}}
{{end-eqn}}
{{qed}}