The necessary conditions for a general second degree equation

For the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ to represent a genuine pair of straight lines, two conditions must both hold:

1. $abc+2fgh-af^2-bg^2-ch^2=0$ (the determinant condition), and
2. $h^2-ab \geq 0$. Condition (i) is exactly the vanishing of the $3\times3$ determinant $\begin{vmatrix}a&h&g\\h&b&f\\g&f&c\end{vmatrix}$, whose expansion along the first row reproduces $abc+2fgh-af^2-bg^2-ch^2$. Condition (i) alone is not sufficient on its own to prove a pair of lines -- both conditions are needed together, and in practice the safest method for identifying the separate lines is to factor the quadratic part and then find the two linear constants by comparison, as in Example 1 below, rather than relying only on the determinant formula. Whenever an equation does represent a pair of lines, several useful facts follow automatically:

1. The two lines are parallel to the lines represented by the homogeneous part $ax^2+2hxy+by^2=0$ (dropping the general equation's linear and constant terms leaves the equation of the pair of lines through the origin parallel to the original pair).

2. The acute angle between them is still given by $\tan\theta = \dfrac{2\sqrt{h^2-ab}}{a+b}$, exactly as for the homogeneous case, since translating a line does not change its slope.

3. The lines are perpendicular exactly when $a+b=0$.

4. The lines are parallel to each other (coincident direction) exactly when $h^2-ab=0$.

5. When $h^2-ab>0$ the two lines genuinely intersect, at the point $\left(\dfrac{hf-bg}{ab-h^2},\ \dfrac{gh-af}{ab-h^2}\right)$.

6. The joint equation of the two bisectors of the angle between the lines represented by $ax^2+2hxy+by^2=0$ is $hx^2-(a-b)xy-hy^2=0$; since the coefficient of $x^2$ plus the coefficient of $y^2$ in this bisector equation is $h+(-h)=0$, the two bisectors are always perpendicular to each other -- exactly as the two angle bisectors of any pair of lines should be.

Worked Examples

Example 1. Show that $x^2-6xy+5y^2+10x-14y+9=0$ represents a pair of lines; find the acute angle between them and their point of intersection.

The quadratic part factors as $x^2-6xy+5y^2=(x-5y)(x-y)$. Try writing the full equation as $(x-5y+c)(x-y+k)$ for constants $c,k$: expanding gives $x^2-6xy+5y^2+(c+k)x-(c+5k)y+ck$. Comparing with the given linear and constant terms: $c+k=10$, $c+5k=14$, $ck=9$. Testing $c=9,k=1$ satisfies all three (check: $9+1=10$ ✓, $9+5=14$ ✓, $9\times1=9$ ✓), so the equation factors as $(x-5y+9)(x-y+1)=0$, confirming it represents a pair of intersecting lines.

The acute angle: with $a=1,h=-3,b=5$, $\tan\theta=\dfrac{2\sqrt{9-5}}{1+5}=\dfrac{2(2)}{6}=\dfrac23$, so $\theta=\tan^{-1}\dfrac23$.

The point of intersection is where both factors vanish: $x-5y+9=0$ and $x-y+1=0$. Subtracting, $-4y+8=0 \Rightarrow y=2$, then $x=y-1=1$. So the lines meet at $(1,2)$ (matching the general point-of-intersection formula with $g=5,f=-7,c=9,a=1,h=-3,b=5$).

Example 2. Find $k$ if $2x^2+4xy-2y^2+4x+8y+k=0$ represents a pair of lines.

Here $a=2,b=-2,c=k,f=4,g=2,h=2$. The determinant condition gives $abc+2fgh-af^2-bg^2-ch^2=0$: $(2)(-2)(k)+2(4)(2)(2)-2(16)-(-2)(4)-(k)(4)=0$, i.e. $-4k+32-32+8-4k=0$, i.e. $-8k+8=0$, so $k=1$.

Example 3. Find $p$ and $q$ if $2x^2+4xy-py^2+4x+qy+1=0$ represents a pair of perpendicular lines.

Here $a=2,b=-p,c=1,f=\dfrac{q}{2},g=2,h=2$. Perpendicularity requires $a+b=0$: $2-p=0 \Rightarrow p=2$. The determinant condition $abc+2fgh-af^2-bg^2-ch^2=0$ becomes $(2)(-p)(1)+2\left(\dfrac q2\right)(2)(2)-2\left(\dfrac q2\right)^2-(-p)(4)-1(4)=0$, i.e. $-2p+4q-\dfrac{q^2}2+4p-4=0$, i.e. $2p+4q-\dfrac{q^2}2-4=0$. Substituting $p=2$: $4+4q-\dfrac{q^2}2-4=0 \Rightarrow 4q-\dfrac{q^2}2=0 \Rightarrow 8q-q^2=0 \Rightarrow q(8-q)=0$, so $q=0$ or $q=8$. …