Parallel Pair of Straight Lines

Condition Parallel Pair of Straight Lines

Consider a pair of straight lines given by the equation:

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

Writing it as a quadratic equation in \(x\):

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

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

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

Simplifying under the square root:

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

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

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

There is a possibility that \(h^2 - ab = 0\) and \(hg - af = 0\) (while \(g^2 - ac\) may not be zero). In that case:

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

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

\[ x = \frac{-(hy + g) + \sqrt{g^2 - ac}}{a} \quad \text{and} \quad x = \frac{-(hy + g) - \sqrt{g^2 - ac}}{a} \]

These correspond to the equations:

\[ ax + hy + g + \sqrt{g^2 - ac} = 0 \]

\[ ax + hy + g - \sqrt{g^2 - ac} = 0 \]

Therefore, we get two parallel straight lines:

\[ ax + hy + g + \sqrt{g^2 - ac} = 0 \]

\[ ax + hy + g - \sqrt{g^2 - ac} = 0 \]

Thus, if \(h^2 - ab = 0\) and \(hg - af = 0\), the original quadratic equation represents two parallel straight lines.

\[ h^2 - ab = 0 \quad \& \quad hg - af = 0 \]

\[ \implies \frac{h}{a} = \frac{b}{h} \quad \& \quad \frac{h}{a} = \frac{f}{g} \]

\[ \implies \frac{b}{h} = \frac{f}{g} \]

\[ \implies hf - bg = 0 \]

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

\[ = (abc - ah^2) + (fgh - af^2) + (fgh - bg^2) \]

\[ = c(ab - h^2) + f(hg - af) + g(hf - bg) \]

\[ = c \cdot 0 + f \cdot 0 + g \cdot 0 = 0 \]

So, the equation

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

represents a parallel pair of straight lines if

\[ \frac{h}{a} = \frac{b}{h} = \frac{f}{g} \]

Coincident Lines

If \(g^2 - ac = 0\) (which will also mean \(f^2 - bc = 0\)), then the lines are coincident.

When \(g^2 - ac = 0\), the equation for the lines simplifies as follows:

Given:

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

We previously found:

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

If \(g^2 - ac = 0\), the square root term vanishes:

\[ x = \frac{-(hy + g)}{a} \]

This means both solutions for \(x\) are the same, indicating coincident lines. Specifically:

\[ ax + hy + g = 0 \]

Similarly, considering the equation as a quadratic in \(y\):

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

We use the quadratic formula to solve for \(y\):

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

Simplifying under the square root:

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

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

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

For the expression inside the square root to be zero:

\[ (h^2 - ab)x^2 + 2(hf - ag)x + (f^2 - ac) = 0 \]

If \(h^2 - ab = 0\) and \(hf - ag = 0\), and also \(f^2 - ac = 0\), we get coincident lines.

So, the conditions for coincident lines are:

\[ h^2 = ab \]

\[ hg = af \]

\[ g^2 = ac \]

\[ f^2 = bc \]

Under these conditions, the equation:

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

represents coincident lines. These conditions ensure that the quadratic equation simplifies into a single pair of overlapping lines, confirming that the lines are coincident.

The condition \( h^2 - ab = 0 \) indicates that the first three terms of the quadratic equation \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) form a perfect square. Specifically, this can be written as:

\[ ( \sqrt{a} x + \sqrt{b} y )^2 + 2gx + 2fy + c = 0 \]

When \( h^2 - ab = 0 \), it means:

\[ h = \sqrt{ab} \]

So the expression \( ax^2 + 2hxy + by^2 \) becomes:

\[ ax^2 + 2\sqrt{ab}xy + by^2 = (\sqrt{a}x + \sqrt{b}y)^2 \]

Therefore, the quadratic equation can be rewritten as:

\[ (\sqrt{a}x + \sqrt{b}y)^2 + 2gx + 2fy + c = 0 \]

Distance between a parallel pair of straight lines

Consider a parallel pair of straight lines represented by the quadratic equation:

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

Writing it as a quadratic equation in \(x\):

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

For the lines to be parallel, we set \(h^2 - ab = 0\) and \(hf - bg = 0\). Under these conditions, we get:

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

This simplifies to the equations of two parallel lines:

\[ ax + hy + g + \sqrt{g^2 - ac} = 0 \]

\[ ax + hy + g - \sqrt{g^2 - ac} = 0 \]

The distance \(d\) between these two parallel lines can be calculated using the formula for the distance between two parallel lines \(ax + by + c_1 = 0\) and \(ax + by + c_2 = 0\):

\[ d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \]

In this case, \(c_1 = g + \sqrt{g^2 - ac}\) and \(c_2 = g - \sqrt{g^2 - ac}\). Therefore, the distance \(d\) between the two lines is:

\[ d = \frac{|(g + \sqrt{g^2 - ac}) - (g - \sqrt{g^2 - ac})|}{\sqrt{a^2 + h^2}} \]

\[ d = \frac{|g + \sqrt{g^2 - ac} - g + \sqrt{g^2 - ac}|}{\sqrt{a^2 + h^2}} \]

\[ d = \frac{|2\sqrt{g^2 - ac}|}{\sqrt{a^2 + h^2}} \]

Substituting \(h^2 = ab\) in the denominator:

\[ d = \frac{2\sqrt{g^2 - ac}}{\sqrt{a^2 + ab}} \]

\[ d = \frac{2\sqrt{g^2 - ac}}{\sqrt{a(a + b)}} \]

Thus, the distance between the two parallel lines is:

\[ \boxed{d = \frac{2\sqrt{g^2 - ac}}{\sqrt{a(a + b)}}} \]

Alternatively,

Writing the given pair as a quadratic equation in \(y\):

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

For the lines to be parallel, we set \(h^2 - ab = 0\) and \(hf - bg = 0\). Under these conditions, we get:

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

This simplifies to the equations of two parallel lines:

\[ bx + hy + f + \sqrt{f^2 - bc} = 0 \]

\[ bx + hy + f - \sqrt{f^2 - bc} = 0 \]

The distance \(d\) between these two parallel lines can be calculated using the formula for the distance between two parallel lines \(ax + by + c_1 = 0\) and \(ax + by + c_2 = 0\):

\[ d = \frac{|(f + \sqrt{f^2 - bc}) - (f - \sqrt{f^2 - bc})|}{\sqrt{a^2 + b^2}} \]

\[ d = \frac{|2\sqrt{f^2 - bc}|}{\sqrt{a^2 + b^2}} \]

Substituting \(h^2 = ab\) in the denominator:

\[ d = \frac{2\sqrt{f^2 - bc}}{\sqrt{b^2 + ab}} \]

\[ d = \frac{2\sqrt{f^2 - bc}}{\sqrt{b(a + b)}} \]

Thus, the distance between the two parallel lines is:

\[ \boxed{d = \frac{2\sqrt{f^2 - bc}}{\sqrt{b(a + b)}}} \]

The distance between the parallel straight lines represented by the quadratic equation \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) is:

\[ d = \frac{2\sqrt{g^2 - ac}}{\sqrt{a(a + b)}} \]

or

\[ d = \frac{2\sqrt{f^2 - bc}}{\sqrt{b(a + b)}} \]