The combined equation of a pair of straight lines L_{1} ≡ a_{1}x + b_{1}y + c_{1} = 0 and L_{2} ≡ a_{2}x + b_{2}y + c_{2} = 0 is (a_{1}x + b_{1}y + c_{1}) (a_{2}x + b_{2}y + c_{2}) = 0 i.e. L_{1} L_{2} = 0.

Opening the brackets and comparing the terms with the terms of general equation of 2nd degree ax^{2} + 2hxy + by^{2} + 2gx +2fy + c= 0, we can get all the following results for pair of straight lines.

The general equation of second degree ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represents a pair of straight lines if :

$ \large \left| \begin{array}{ccc} a & h & g \\ g & f & c \\ h & b & f \end{array} \right| = 0 $ and $\large h^2 \ge a b$

abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0 and h^{2} ≥ ab.

The homogeneous second degree equation ax^{2} + 2hxy + by^{2} = 0 represents a pair of straight lines through the origin if h^{2} ≥ ab.

If the lines through the origin whose joint equation is ax^{2} + 2hxy + by^{2} = 0, are y = m_{1}x and

y = m_{2}x, then y^{2} – (m_{1} + m_{2})xy + m_{1}m_{2}x^{2} = 0 and $\large y^2 + \frac{2h}{b}xy + \frac{a}{b}x^2 = 0 $

are identical, so that $\large m_1 + m_2 = -\frac{2h}{b} , m_1 m_2 = \frac{a}{b} $

If θ be the angle between two lines, through the origin, then

$\large tan\theta = \pm \frac{\sqrt{(m_1 + m_2)^2 – 4 m_1 m_2}}{1 + m_1 m_2} $

$\large tan\theta = \pm \frac{2\sqrt{h^2 -ab}}{a+b} $

The lines are perpendicular if a + b = 0 and coincident if h^{2} = ab.

In the more general case, the lines represented by ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 will be perpendicular if a + b = 0, parallel if the terms of second degree make a perfect square i.e.

ax^{2} +2hxy + by^{2} gets converted into (l_{1}x ± m_{1}y)^{2}, coincident if the whole equation makes a perfect square i.e. ax^{2} + 2hxy +by^{2} + 2gx +2fy + c can be written as (lx + my + n)^{2 }.

**Note:**

- Point of intersection of the two lines represented by ax
^{2}+2hxy+by^{2}+2gx+2fy+c=0 is obtained by solving the equations ∂f/∂x = ax + hy + g = 0 and ∂f/∂y = hx + by + f = 0 where ∂f/∂x denotes the derivative of f with respect to x, keeping y constant and ∂f/∂y denotes the derivative of f with respect to y, keeping x constant. The fact can be used in splitting ax^{2}+ by^{2}+ 2hxy + 2gx + 2fy + c = 0 into equations of two straight lines. With the above method, the point of intersection can be found. Now only the slopes need to be determined.

It should be noted that the line ax + hy + g = 0 and hx + by + f = 0 are not the lines represented by ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0. These are the lines concurrent with the lines represented by given equation.