General Pair of Straight Lines

General Pair of Straight Lines

Consider two straight lines given by the equations:

\[ l_1 x + m_1 y + n_1 = 0 \]

and

\[ l_2 x + m_2 y + n_2 = 0 \]

The joint equation representing these two lines can be obtained by multiplying their individual equations:

\[ (l_1 x + m_1 y + n_1)(l_2 x + m_2 y + n_2) = 0 \]

Expanding this product, we get:

\[ l_1 l_2 x^2 + (l_1 m_2 + l_2 m_1) xy + m_1 m_2 y^2 + (l_1 n_2 + l_2 n_1) x + (m_1 n_2 + m_2 n_1) y + n_1 n_2 = 0 \]

This expanded form has the general quadratic structure:

\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]

where: - \( a = l_1 l_2 \) - \( 2h = l_1 m_2 + l_2 m_1 \) - \( b = m_1 m_2 \) - \( 2g = l_1 n_2 + l_2 n_1 \) - \( 2f = m_1 n_2 + m_2 n_1 \) - \( c = n_1 n_2 \)

Converse Analysis

A natural question arises: does the general second-degree equation

\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]

always represent a pair of straight lines?

The answer is no. The quadratic equation represents a pair of straight lines only under specific conditions. This is what we will first analyze: determining the conditions under which the equation represents a pair of straight lines.

Factorization

To answer this question, let us first consider an example.

Take the lines \(2x + y - 3 = 0\) and \(x + 3y - 1 = 0\). To find their joint equation, we multiply the individual equations:

\[ (2x + y - 3)(x + 3y - 1) = 0 \]

Expanding this product, we get:

\[ (2x + y - 3)(x + 3y - 1) = 2x(x + 3y - 1) + y(x + 3y - 1) - 3(x + 3y - 1) \]

\[ = 2x^2 + 6xy - 2x + xy + 3y^2 - y - 3x - 9y + 3 \]

\[ = 2x^2 + 7xy + 3y^2 - 5x - 10y + 3 = 0 \]

This expanded form is the joint equation of the lines \(2x + y - 3 = 0\) and \(x + 3y - 1 = 0\):

\[ 2x^2 + 7xy + 3y^2 - 5x - 10y + 3 = 0 \]

The next step is to ask how one can get back the original equations from this expanded quadratic form. In other words, how can we factorize the quadratic equation to retrieve the two linear factors?

Given the quadratic equation:

\[ 2x^2 + 7xy + 3y^2 - 5x - 10y + 3 = 0 \]

To factorize the given example equation \(2x^2 + 7xy + 3y^2 - 5x - 10y + 3 = 0\), we express it as a quadratic equation in \(x\):

\[ 2x^2 + (7y - 5)x + (3y^2 - 10y + 3) = 0 \]

Next, we solve for \(x\) using the quadratic formula:

\[ x = \frac{-(7y - 5) \pm \sqrt{(7y - 5)^2 - 4 \cdot 2 \cdot (3y^2 - 10y + 3)}}{2 \cdot 2} \]

\[ x = \frac{-(7y - 5) \pm \sqrt{49y^2 - 70y + 25 - 24y^2 + 80y - 24}}{4} \]

\[ x = \frac{-(7y - 5) \pm \sqrt{25y^2 + 10y + 1}}{4} \]

\[ x = \frac{-(7y - 5) \pm \sqrt{(5y + 1)^2}}{4} \]

\[ x = \frac{-(7y - 5) \pm (5y + 1)}{4} \]

This gives us two solutions for \(x\):

\[ x = \frac{-(7y - 5) + (5y + 1)}{4} = \frac{-7y + 5 + 5y + 1}{4} = \frac{-2y + 6}{4} = \frac{-y + 3}{2} \]

\[ x = \frac{-(7y - 5) - (5y + 1)}{4} = \frac{-7y + 5 - 5y - 1}{4} = \frac{-12y + 4}{4} = -3y + 1 \]

Thus, the two factors of the quadratic equation are:

\[ x = \frac{-y + 3}{2} \implies 2x + y - 3 = 0 \]

\[ x = -3y + 1 \implies x + 3y - 1 = 0 \]

Therefore, we have successfully factorized the quadratic equation back into its original linear equations:

\[ 2x + y - 3 = 0 \]

\[ x + 3y - 1 = 0 \]

This process illustrates how to factorize a quadratic equation representing a pair of straight lines by expressing it as a quadratic equation in one variable and then using the quadratic formula to find the linear factors.

Let us apply the same process to a general quadratic equation representing a pair of straight lines, consider the equation:

\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]

We express this equation as a quadratic in \(x\):

\[ ax^2 + 2(hy + g)x + (by^2 + 2fy + c) = 0 \]

Using the quadratic formula to solve for \(x\):

\[ x = \frac{-2(hy + g) \pm \sqrt{[2(hy + g)]^2 - 4a(by^2 + 2fy + c)}}{2a} \]

Simplifying the discriminant:

\[ x = \frac{-2(hy + g) \pm \sqrt{4(hy + g)^2 - 4a(by^2 + 2fy + c)}}{2a} \]

\[ x = \frac{-2(hy + g) \pm \sqrt{4(h^2y^2 + 2hgy + g^2 - aby^2 - 2afy - ac)}}{2a} \]

\[ x = \frac{-2(hy + g) \pm 2\sqrt{(h^2 - ab)y^2 + 2(hg - af)y + g^2 - ac}}{2a} \]

\[ x = \frac{-(hy + g) \pm \sqrt{(h^2 - ab)y^2 + 2(hg - af)y + g^2 - ac}}{a} \]

For this quadratic equation to represent a pair of straight lines, the expression inside the square root must be a perfect square. This requires that the discriminant of the quadratic expression inside the square root, which is a quadratic equation in \(y\), should be zero.

The discriminant of the quadratic expression \((h^2 - ab)y^2 + 2(hg - af)y + (g^2 - ac)\) is:

\[ \Delta = [2(hg - af)]^2 - 4(h^2 - ab)(g^2 - ac) \]

\[ \Delta = 4(hg - af)^2 - 4(h^2 - ab)(g^2 - ac) \]

\[ \Delta = 4[(hg - af)^2 - (h^2 - ab)(g^2 - ac)] \]

For this discriminant to be zero:

\[ (hg - af)^2 - (h^2 - ab)(g^2 - ac) = 0 \]

Expanding and simplifying, we get the condition:

\[ (hg - af)^2 - (h^2g^2 - ac h^2 - ab g^2 + ab ac) = 0 \]

Simplifying further, we obtain:

\[ (hg - af)^2 - h^2g^2 + h^2ac + abg^2 - abac = 0 \]

Which simplifies to:

\[ abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \]

Thus, for the quadratic equation \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) to represent a pair of straight lines, it must satisfy the condition:

\[ abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \]

Furthermore, the term under the square root \((h^2 - ab)y^2 + 2(hg - af)y + (g^2 - ac)\) must be non-negative for real values of \(x\). This implies that:

\[ h^2 - ab \geq 0 \]

If \(h^2 - ab < 0\), the expression inside the square root is always negative, indicating that the original quadratic equation does not represent a pair of real straight lines.

Therefore, the quadratic equation \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) represents a pair of straight lines if and only if:

1. \( h^2 - ab \geq 0 \)
2. \( abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \)

When these conditions are met, the quadratic equation can be factored into two linear equations, confirming that it represents a pair of straight lines.

A quadratic expression \(ax^2 + bx + c\) for which \(a > 0\) and \(b^2 - 4ac = 0\) implies that \(c\) is always \(\geq 0\). This can be understood from the graph of the quadratic expression.

When \(a > 0\) and \(b^2 - 4ac = 0\), the graph is a parabola that opens upwards and touches the x-axis at a single point, indicating that the vertex of the parabola is on the x-axis. Consequently, the y-intercept, which is \(c\), cannot be negative. Therefore, \(c \geq 0\).

Applying this understanding to the quadratic expression inside the square root in our general equation, we can conclude that if the expression represents a pair of straight lines (i.e., both conditions 1 and 2 are satisfied), then:

\[ g^2 - ac \geq 0 \]

This follows because the expression inside the square root, \((h^2 - ab)y^2 + 2(hg - af)y + (g^2 - ac)\), must be a non-negative perfect square. Given that \(h^2 - ab \geq 0\), the quadratic term in \(y\) must be non-negative. Consequently, the constant term \(g^2 - ac \geq 0\).

Thus, the condition \(g^2 - ac \geq 0\) ensures that the quadratic expression inside the square root is a non-negative perfect square, which is necessary for the original quadratic equation \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) to represent a pair of straight lines.

We can also write the quadratic equation \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) as a quadratic equation in terms of \(y\):

\[ by^2 + 2(hx + f)y + (ax^2 + 2gx + c) = 0 \]

Using the quadratic formula to solve for \(y\):

\[ y = \frac{-(2hx + 2f) \pm \sqrt{[2(hx + f)]^2 - 4b(ax^2 + 2gx + c)}}{2b} \]

Simplifying the discriminant:

\[ y = \frac{-(2hx + 2f) \pm \sqrt{4(hx + f)^2 - 4b(ax^2 + 2gx + c)}}{2b} \]

\[ y = \frac{-(2hx + 2f) \pm \sqrt{4(h^2x^2 + 2hfx + f^2 - abx^2 - 4bgx - 4bc)}}{2b} \]

\[ y = \frac{-(2hx + 2f) \pm 2\sqrt{(h^2 - ab)x^2 + 2(hf - bg)x + f^2 - bc}}{2b} \]

\[ y = \frac{-(hx + f) \pm \sqrt{(h^2 - ab)x^2 + 2(hf - bg)x + f^2 - bc}}{b} \]

For this quadratic equation to represent a pair of straight lines, the expression inside the square root must be a perfect square. This requires that the discriminant of the quadratic expression inside the square root, which is a quadratic equation in \(x\), should be zero.

The discriminant of the quadratic expression \((h^2 - ab)x^2 + 2(hf - bg)x + (f^2 - bc)\) is:

\[ \Delta = [2(hf - bg)]^2 - 4(h^2 - ab)(f^2 - bc) \]

\[ \Delta = 4(hf - bg)^2 - 4(h^2 - ab)(f^2 - bc) \]

\[ \Delta = 4[(hf - bg)^2 - (h^2 - ab)(f^2 - bc)] \]

For this discriminant to be zero:

\[ (hf - bg)^2 - (h^2 - ab)(f^2 - bc) = 0 \]

Expanding and simplifying, we get the condition:

\[ (hf - bg)^2 - (h^2f^2 - abf^2 + h^2bc - abbc) = 0 \]

Simplifying further, we obtain:

\[ (hf - bg)^2 - h^2f^2 + abf^2 + h^2bc - abbc = 0 \]

Which simplifies to:

\[ abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \]

Therefore, for the quadratic equation \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) to represent a pair of straight lines, it must satisfy the condition:

\[ abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \]

Additionally, we can observe that for the expression to be a non-negative perfect square, it must meet several conditions. When written as a quadratic in \(x\) and \(y\), both the constant and quadratic terms must be non-negative.

Therefore:

1. \( h^2 - ab \geq 0 \)
2. \( f^2 - bc \geq 0 \)
3. \( g^2 - ac \geq 0 \)

Thus, for the quadratic equation \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) to represent a pair of straight lines, it must satisfy the following conditions:

1. \( h^2 - ab \geq 0 \)
2. \( f^2 - bc \geq 0 \)
3. \( g^2 - ac \geq 0 \)
4. \( abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \)

Of which conditions 1 and 4 together are necessary and sufficient. If 1 and 4 are true, then 2 and 3 can never be false. If any of the condition is false, then the equation provided is not a pair of straight lines.

\[ abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \]

This condition can be represented in determinant form as:

\[ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \]