Some Important Results on Pair of Straight Lines

Product of Perpendiculars

Finding the Product of Perpendiculars from a Point to a Pair of Straight Lines

Consider a pair of straight lines given by the equation:

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

We wish to find the product of the perpendiculars drawn from the point \((\alpha, \beta)\) to this pair of straight lines.

Previously, we saw that the product of the perpendiculars from \((\alpha, \beta)\) to the pair of lines represented by \( ax^2 + 2hxy + by^2 = 0 \) is given by:

\[ \frac{|a\alpha^2 + 2h\alpha\beta + b\beta^2|}{\sqrt{(a - b)^2 + 4h^2}} \]

Let \((x_1, y_1)\) be the intersection point of the general pair of straight lines. Shift the origin to \((x_1, y_1)\). Then, the coordinates \((\alpha, \beta)\) transform to \((\alpha - x_1, \beta - y_1)\), and the general pair of straight lines transforms to:

\[ ax^2 + 2hxy + by^2 = 0 \]

The product of the perpendiculars from \((\alpha - x_1, \beta - y_1)\) to the transformed pair of lines is:

\[ \frac{|a(\alpha - x_1)^2 + 2h(\alpha - x_1)(\beta - y_1) + b(\beta - y_1)^2|}{\sqrt{(a - b)^2 + 4h^2}} \]

Expanding the numerator:

\[ a(\alpha - x_1)^2 + 2h(\alpha - x_1)(\beta - y_1) + b(\beta - y_1)^2 \]

Expanding the terms:

\[ a(\alpha^2 - 2\alpha x_1 + x_1^2) + 2h(\alpha\beta - \alpha y_1 - \beta x_1 + x_1 y_1) + b(\beta^2 - 2\beta y_1 + y_1^2) \]

Combining like terms, we get:

\[ a\alpha^2 - 2a\alpha x_1 + ax_1^2 + 2h\alpha\beta - 2h\alpha y_1 - 2h\beta x_1 + 2hx_1 y_1 + b\beta^2 - 2b\beta y_1 + by_1^2 \]

Rewriting, we observe:

\[ a\alpha^2 + 2h\alpha\beta + b\beta^2 - 2\alpha(ax_1 + hy_1) - 2\beta(hx_1 + by_1) + ax_1^2 + 2hx_1 y_1 + by_1^2 \]

We know that:

\[ ax_1 + hy_1 + g = 0 \quad (1)\]

\[ hx_1 + by_1 + f = 0 \quad (2)\]

\[ gx_1 + fy_1 + c = 0 \quad (3) \]

Since \((x_1, y_1)\) lies on the pair of straight lines:

\[ ax_1^2 + 2hx_1 y_1 + by_1^2 + 2gx_1 + 2fy_1 + c = 0 \]

\[ ax_1^2 + 2hx_1 y_1 + by_1^2 + 2(gx_1 + fy_1) + c = 0 \]

Using these equation (3) , we get:

\[ ax_1^2 + 2hx_1 y_1 + by_1^2 - 2c + c = 0 \]

\[ ax_1^2 + 2hx_1 y_1 + by_1^2 = c \quad (4)\]

Using (1) and (2) and (4) the expanded numerator simplifies to:

\[ a\alpha^2 + 2h\alpha\beta + b\beta^2 + 2g\alpha + 2f\beta + c \]

The product of the perpendiculars is:

\[ \boxed{\frac{|a\alpha^2 + 2h\alpha\beta + b\beta^2 + 2g\alpha + 2f\beta + c|}{\sqrt{(a - b)^2 + 4h^2}} }\]

Product of Perpendiculars from origin

Using the formula derived earlier, the product of perpendiculars from the origin to the pair of straight lines is given by:

\[ \frac{|a\alpha^2 + 2h\alpha\beta + b\beta^2 + 2g\alpha + 2f\beta + c|}{\sqrt{(a - b)^2 + 4h^2}} \]

Substituting \(\alpha = 0\) and \(\beta = 0\):

\[ \frac{|a(0)^2 + 2h(0)(0) + b(0)^2 + 2g(0) + 2f(0) + c|}{\sqrt{(a - b)^2 + 4h^2}} \]

This simplifies to:

\[ \frac{|c|}{\sqrt{(a - b)^2 + 4h^2}} \]

Therefore, the product of perpendiculars from the origin to the pair of straight lines represented by the equation \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) is:

\[ \frac{|c|}{\sqrt{(a - b)^2 + 4h^2}} \]

A pair of lines equidistant from origin

If the pair of straight lines \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) is such that both lines in the pair are at the same distance from the origin, then \( f^4 - g^4 = c(bf^2 - ag^2) \).

Proof:

Let \( y = m_1 x + c_1 \) and \( y = m_2 x + c_2 \) be the lines represented by the given pair.

We know the following relationships from the coefficients of the quadratic equation:

\[ m_1 + m_2 = -\frac{2h}{b} \]

\[ m_1 m_2 = \frac{a}{b} \]

\[ m_1 c_2 + m_2 c_1 = \frac{2g}{b} \]

\[ c_1 + c_2 = -\frac{2f}{b} \]

\[ c_1 c_2 = \frac{c}{b} \]

The distance of the origin from the line \( y = m_1 x + c_1 \) is equal to the distance from the line \( y = m_2 x + c_2 \). Thus:

\[ \frac{|c_1|}{\sqrt{1 + m_1^2}} = \frac{|c_2|}{\sqrt{1 + m_2^2}} \]

Squaring both sides and simplifying, we get:

\[ c_1^2 (1 + m_2^2) = c_2^2 (1 + m_1^2) \]

\[ c_1^2 - c_2^2 = m_2^2 c_2^2 - m_1^2 c_1^2 \]

\[ (c_1 + c_2)(c_1 - c_2) = (m_1 c_2 + m_2 c_1)(m_1 c_2 - m_2 c_1) \]

\[ (c_1 + c_2)^2 (c_1 - c_2)^2 = (m_1 c_2 + m_2 c_1)^2 (m_1 c_2 - m_2 c_1)^2 \]

\[ \Rightarrow (c_1 + c_2)^2 (c_1 - c_2)^2 = (m_1 c_2 + m_2 c_1)^2 (m_1 c_2 - m_2 c_1)^2 \]

\[ \Rightarrow (c_1 + c_2)^2 \left[(c_1 + c_2)^2 - 4c_1 c_2 \right] = (m_1 c_2 + m_2 c_1)^2 \left[(m_1 c_2 + m_2 c_1)^2 - 4m_1 m_2 c_1 c_2 \right] \]

Substituting the known values from the quadratic coefficients:

\[ \left(\frac{4f^2}{b^2}\right) = \left(\frac{4g^2}{b^2} - \frac{4ac}{b^2}\right) \]

Simplifying further, we get:

\[ f^2(f^2 - c^2) = g^2(g^2 - ac) \]

\[ \implies f^4 - f^2 c^2 = g^4 - g^2 ac \]

\[ \implies f^4 - g^4 = c(bf^2 - ag^2) \]

Hence, it follows that:

\[ f^4 - g^4 = c(bf^2 - ag^2) \]

This completes the proof of the theorem.

Area of Triangle

Consider the area of a triangle formed by the line \(lx + my + n = 0\) with a pair of straight lines \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\), having \((x_1, y_1)\) as the point of intersection.

The area of the triangle is given by:

\[ \text{Area} = \frac{(lx_1 + my_1 + n)^2 \sqrt{h^2 - ab}}{|am^2 - 2hlm + bl^2|} \]

Proof:

We know that the area of the triangle formed by a line \(lx + my + n = 0\) with the pair of straight lines \(ax^2 + 2hxy + by^2 = 0\) is:

\[ \text{Area} = \frac{n^2 \sqrt{h^2 - ab}}{|am^2 - 2hlm + bl^2|} \]

To generalize this for the pair of straight lines intersecting at \((x_1, y_1)\), we shift the origin to \((x_1, y_1)\):

\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \quad \text{transforms to} \quad ax^2 + 2hxy + by^2 = 0 \]

The line \(lx + my + n\) transforms as follows:

\[ l(x + x_1) + m(y + y_1) + n = 0 \implies lx + my + (lx_1 + my_1 + n) = 0 \]

Therefore, the area of the triangle formed is:

\[ \text{Area} = \frac{(lx_1 + my_1 + n)^2 \sqrt{h^2 - ab}}{|am^2 - 2hlm + bl^2|} \]

Intercepts on Axes

Length of x-intercept

The length of the intercept made by a general pair of straight lines \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) on the x-axis is \( \frac{2\sqrt{g^2 - ac}}{|a|} \).

Proof: Consider the given equation of the pair of straight lines \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \).

When these lines intersect the x-axis, \( y = 0 \). Substituting \( y = 0 \) in the equation, we obtain:

\[ ax^2 + 2gx + c = 0 \]

This is a quadratic equation in \( x \).

Let \( x_1 \) and \( x_2 \) be the roots of the quadratic equation. According to Vieta's formulas:

\[ x_1 + x_2 = -\frac{2g}{a} \quad \text{and} \quad x_1 x_2 = \frac{c}{a} \]

The length of the intercept \( AB \) on the x-axis is given by the absolute difference of the roots \( x_1 \) and \( x_2 \):

\[ |x_1 - x_2| = \sqrt{(x_1 + x_2)^2 - 4x_1 x_2} \]

Substituting the values from Vieta's formulas:

\[ |x_1 - x_2| = \sqrt{\left( -\frac{2g}{a} \right)^2 - 4 \left( \frac{c}{a} \right)} \]

\[ = \sqrt{\frac{4g^2}{a^2} - \frac{4ac}{a^2}} \]

\[ = \sqrt{\frac{4(g^2 - ac)}{a^2}} \]

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

Therefore, the length of the intercept made by the pair of straight lines on the x-axis is:

\[ \boxed{\frac{2\sqrt{g^2 - ac}}{|a|}} \]

Length of y-intercept

The length of the intercept made by a general pair of straight lines \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) on the y-axis is \( \frac{2\sqrt{f^2 - bc}}{|b|} \).

Proof:

Consider the given equation of the pair of straight lines \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \).

When these lines intersect the y-axis, \( x = 0 \). Substituting \( x = 0 \) in the equation, we obtain:

\[ by^2 + 2fy + c = 0 \]

This is a quadratic equation in \( y \).

Let \( y_1 \) and \( y_2 \) be the roots of the quadratic equation. According to Vieta's formulas:

\[ y_1 + y_2 = -\frac{2f}{b} \quad \text{and} \quad y_1 y_2 = \frac{c}{b} \]

The length of the intercept \( CD \) on the y-axis is given by the absolute difference of the roots \( y_1 \) and \( y_2 \):

\[ |y_1 - y_2| = \sqrt{(y_1 + y_2)^2 - 4y_1 y_2} \]

Substituting the values from Vieta's formulas:

\[ |y_1 - y_2| = \sqrt{\left( -\frac{2f}{b} \right)^2 - 4 \left( \frac{c}{b} \right)} \]

\[ = \sqrt{\frac{4f^2}{b^2} - \frac{4bc}{b^2}} \]

\[ = \sqrt{\frac{4(f^2 - bc)}{b^2}} \]

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

Therefore, the length of the intercept made by the pair of straight lines on the y-axis is:

\[ \boxed{\frac{2\sqrt{f^2 - bc}}{|b|}} \]

Angle Bisectors

The equation of the pair of bisectors of a pair of straight lines \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) is given by:

\[ h[(x - x_1)^2 - (y - y_1)^2] = (a - b)(x - x_1)(y - y_1) \]

where \((x_1, y_1)\) is the intersection point of the given pair of straight lines.

We can prove this easily using shifting the origin method.

Quadrilaterals