Homogenous Pair of Straight Lines
Homogeneous Pair of Straight Lines
A homogeneous pair of straight lines is the joint equation of two straight lines passing through the origin. These lines can be represented by the equations:
The joint equation, or pair of straight lines, is given by:
This represents all points \((x, y)\) that lie on either of the two lines passing through the origin.
For Example,
Consider the lines:
The pair of straight lines is:
Expanding this product, we get:
This quadratic polynomial represents the pair of straight lines passing through the origin. The reason we call it a homogeneous pair is because the final expanded joint equation is a homogeneous polynomial of degree 2. This means that every term in the equation has the same total degree, which is the sum of the exponents of \(x\) and \(y\). In the expanded form \(2x^2 + 7xy  4y^2 = 0\), each term \(2x^2\), \(7xy\), and \(4y^2\) has a total degree of 2. Thus, the equation is homogeneous, and this is why we refer to it as a homogeneous pair of straight lines. By starting with such homogeneous pairs, we lay the groundwork for understanding the more complex general pairs of straight lines.
Consider the following examples:

\[ x(x  y) = 0 \]
This represents the lines \(x = 0\) and \(x  y = 0\) (or \(y = x\)), which intersect at the origin.

\[ (x + y)(x  y) = 0 \]
This represents the lines \(x + y = 0\) and \(x  y = 0\) (or \(y = x\) and \(y = x\)), which intersect at the origin.

\[ (y  3x)(y  6x) = 0 \]
This represents the lines \(y  3x = 0\) (or \(y = 3x\)) and \(y  6x = 0\) (or \(y = 6x\)), which intersect at the origin.

\[ xy = 0 \]
This represents the lines \(x = 0\) and \(y = 0\), which intersect at the origin.
General Form of a Homogenous Pair of Straight Lines
Consider two straight lines passing through the origin:
The corresponding joint equation is:
Expanding this product, we get:
The general form of a pair of straight lines passing through the origin is therefore:
Where:
The reason for writing the coefficient of \(xy\) as \(2h\) is that it simplifies various formulas and calculations later on. By using \(2h\) instead of directly writing the coefficient as \(a_1b_2 + b_1a_2\), we streamline the expressions and derivations involving pairs of straight lines, especially in the context of determinant and discriminant calculations.
All homogeneous pairs of straight lines have the form \( ax^2 + 2hxy + by^2 = 0 \). But does \( ax^2 + 2hxy + by^2 = 0 \) always represent a pair of straight lines? For example, does \( x^2 + 2xy + 6y^2 = 0 \) represent some pair of straight lines?
Determining When \( ax^2 + 2hxy + by^2 = 0 \) Represents a Pair of Straight Lines
Given the general form of a homogeneous pair of straight lines:
We want to determine if this equation represents a pair of straight lines.

Start with the equation:
\[ ax^2 + 2hxy + by^2 = 0 \] 
Rewrite it as:
\[ a x^2 + (2hy)x + by^2 = 0 \] 
Treating this as a quadratic in \( x \):
\[ x = \frac{2hy \pm \sqrt{(2hy)^2  4ab y^2}}{2a} \] 
Simplifying under the square root:
\[ x = \frac{2hy \pm 2y \sqrt{h^2  ab}}{2a} \]\[ x = \frac{hy \pm y \sqrt{h^2  ab}}{a} \] 
Factorizing:
\[ ax = hy \pm y \sqrt{h^2  ab} \]\[ ax + hy = \pm y \sqrt{h^2  ab} \] 
Splitting into two linear equations:
\[ ax + (h + \sqrt{h^2  ab})y = 0 \]\[ ax + (h  \sqrt{h^2  ab})y = 0 \]
Thus, we get two straight lines passing through the origin if \( h^2  ab \ge 0 \).
Therefore, the equation \( ax^2 + 2hxy + by^2 = 0 \) represents the joint equation of two straight lines if \( h^2  ab \ge 0 \). The lines are given by:
This condition ensures that the quadratic equation can be factored into two distinct linear factors, which correspond to two straight lines passing through the origin.
We call \( h^2  ab\) as discriminant of \(ax^2+2hxy+by^2=0\).
Example
Determine if \( x^2 + 2xy + 6y^2 = 0 \) Represents a Pair of Straight Lines
Solution:
To determine if the equation \( x^2 + 2xy + 6y^2 = 0 \) represents a pair of straight lines, we need to check the condition \( h^2  ab \ge 0 \).
Given the equation:
Compare this with the general form:
Here:
Now, calculate the discriminant \( h^2  ab \):
Since the discriminant \( h^2  ab = 5 \) is less than 0, the condition \( h^2  ab \ge 0 \) is not satisfied.
The equation \( x^2 + 2xy + 6y^2 = 0 \) does not represent a pair of straight lines because the discriminant \( h^2  ab \) is less than 0. This indicates that the quadratic polynomial cannot be factored into two distinct linear factors.