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 |
$\blacksquare$
Sources
- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {II}$. The Straight Line: $14$. Condition that the general quadratic equation of the second degree should represent two straight lines