# Characteristic of Quadratic Equation that Represents Two Straight Lines

## Theorem

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 if and only if 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:

 $\ds a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds a^2 x^2 + a b y^2 + 2 a h x y + 2 a g x + 2 a f y + c$ $=$ $\ds 0$ multiplying by $a$ $\ds \leadsto \ \$ $\ds \paren {a x + h y + g}^2 + a b y^2 + 2 a f y + a c$ $=$ $\ds 2 g h y + h^2 y^2 + g^2$ completing the square in $x$ terms $\text {(2)}: \quad$ $\ds \leadsto \ \$ $\ds \paren {a x + h y + g}^2 - \paren {\paren {h^2 - a b} y^2 + 2 \paren {g h - a f} y + \paren {g^2 - a c} }$ $=$ $\ds 0$ rearranging

In order that the second part is a perfect square in $y$, it is necessary that:

 $\ds \paren {g h - a f}^2$ $=$ $\ds \paren {h^2 - a b} \paren {g^2 - a c}$ $\ds \leadsto \ \$ $\ds g^2 h^2 - 2 a f g h + a^2 f^2$ $=$ $\ds g^2 h^2 - a b g^2 - a c h^2 + a^2 b c$ multiplying out $\text {(3)}: \quad$ $\ds \leadsto \ \$ $\ds a b c + 2 f g h - a f^2 - b g^2 - c h^2$ $=$ $\ds 0$ simplifying, rearranging and dividing by $a$ which is non-zero

Conversely, if $(3)$ is true, then $(2)$ can be expressed in the form of a Difference of Two Squares:

 $\text {(2)}: \quad$ $\ds \paren {a x + h y + g}^2 - \paren {\paren {h^2 - a b} y^2 + 2 \paren {g h - a f} y + \paren {g^2 - a c} }$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \paren {a x + h y + g}^2 - \paren {h^2 - a b} \paren {y^2 + 2 \dfrac {g h - a f} {h^2 - a b} y + \dfrac {g^2 - a c} {h^2 - a b} }$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \paren {a x + h y + g}^2 - \paren {h^2 - a b} \paren {\paren {y + \dfrac {g h - a f} {h^2 - a b} }^2 + \dfrac {g^2 - a c} {h^2 - a b} - \paren {\dfrac {g h - a f} {h^2 - a b} }^2}$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \paren {a x + h y + g}^2 - \paren {h^2 - a b} \paren {y + \dfrac {g h - a f} {h^2 - a b} }^2$ $=$ $\ds 0$ as $\paren {g h - a f}^2 - \paren {h^2 - a b} \paren {g^2 - a c}$

Hence $(2)$ has $2$ factors, which can be seen to be the equations of straight lines.

$\Box$

Let $a = 0$ but $b \ne 0$.

Then:

 $\ds b y^2 + 2 h x y + 2 g x + 2 f y + c$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds b^2 y^2 + 2 b h x y + 2 b g x + 2 b f y + b c$ $=$ $\ds 0$ multiplying by $b$ $\ds \leadsto \ \$ $\ds \paren {b y + h x + f}^2 + 2 b g x + b c$ $=$ $\ds 2 f h x + h^2 x^2 + f^2$ completing the square in $y$ terms $\ds \leadsto \ \$ $\ds \paren {b y + h x + f}^2 - \paren {h^2 x^2 + 2 \paren {f h - b g} x + \paren {f^2 - b c} }$ $=$ $\ds 0$ rearranging

In order that the second part is a perfect square in $x$, it is necessary that:

 $\ds \paren {f h - b g}^2$ $=$ $\ds h^2 \paren {f^2 - b c}$ $\ds \leadsto \ \$ $\ds f^2 h^2 - 2 b f g h + b^2 g^2$ $=$ $\ds f^2 h^2 - b c h^2$ multiplying out $\text {(4)}: \quad$ $\ds \leadsto \ \$ $\ds 2 f g h - b g^2 - c h^2$ $=$ $\ds 0$ simplifying, rearranging and dividing by $b$ which is non-zero

it is noted that $(4)$ is the same as $(3)$ but with $a = 0$.

$\Box$

Suppose $a = 0$ and $b = 0$ but $h \ne 0$.

Then:

 $\ds 2 h x y + 2 g x + 2 f y + c$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds 2 h^2 x y + 2 g h x + 2 f h y + c h$ $=$ $\ds 0$ multiplying by $h$ $\text {(5)}: \quad$ $\ds \leadsto \ \$ $\ds 2 \paren {h x + f} \paren {h y + g} + c h$ $=$ $\ds 2 f g$ extracting factors and completing rectangle

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 $b = 0$.

$\Box$

If $a = 0$ and $b = 0$ and $h = 0$, then $(1)$ is not a quadratic equation.

All cases have been covered.

$\Box$

Finally we see that:

 $\ds \begin {vmatrix} a & h & g \\ h & b & f \\ g & f & c \end {vmatrix}$ $=$ $\ds a \begin {vmatrix} b & f \\ f & c \end {vmatrix} - h \begin {vmatrix} h & f \\ g & c \end {vmatrix} + g \begin {vmatrix} h & b \\ g & f \end {vmatrix}$ Determinant of Order 3 $\ds$ $=$ $\ds a \paren {b c - f^2} - h \paren {h c - f g} + g \paren {h f - b g}$ Determinant of Order 2 $\ds$ $=$ $\ds a b c + 2 f g h - a f^2 - b g^2 - c h^2$ simplifying

$\blacksquare$