How to Separate the Equations of Corresponding Straight Lines from a Second-Degree General Equation of Pair of Straight Lines?

Question

I'm trying to understand how to separate the equations of corresponding straight lines from a general second-degree equation representing a pair of straight lines. For example, given the equation:

ax^2+hxy+by^2+gx+fy+c=0

How can I determine the equations of the individual straight lines that make up this pair? I understand that the equation can be factorized into two linear factors, but I'm unsure how to identify and separate the equations of the individual lines.

Could someone provide a step-by-step explanation or method to extract the equations of the corresponding straight lines from such a second-degree equation?

Thank you!

Answer 1

HINTS

Decomposing can be done by evaluating matrix:

$$\{(a,b,g/2),(h,b,f/),(g/2,f/2,c)\}$$

to check that its determinant vanishes.

Next treat

$$ax^2+hxy+by^2+gx+fy+c=0$$

as a quadratic in $y$ solving for two roots $y=f(x)$.

There are two factors, each represents a line (cleaned up if necessary).

I.e., the LHS is decomposable or factorizable into two straight lines.

Answer 2

When $ax^2+hxy+by^2+gx+fy+c=0$ represents two lines, the trick is to find the point $(c_x,c_y)$ where they meet. Writing the equation about that point, the equation reduces to $a(x-c_x)^2+h(x-c_x)(y-c_y)+b(y-c_y)^2=0.$ Then apply the quadratic formula to $at^2+ht+b=0,$ (or $a+hs+bs^2=0$). If the solutions are $t_1,t_2$ making the factorization $a(t-t_1)(t-t_2)=0,$ The original equation becomes $a((x-c_x)-t_1(y-c_y))((x-c_x)-t_2(y-c_y))=0,$ i.e. the lines are $(x-c_x)-t_1(y-c_y)=0$ and $(x-c_x)-t_2(y-c_y)=0.$