Product of Perpendiculars

Suppose \(ax^2 + 2hxy + by^2 = 0\) is a pair of straight lines passing through the origin. Let \((\alpha, \beta)\) be any point in the plane. If \(P_1\) and \(P_2\) are the lengths of the perpendiculars dropped from \((\alpha, \beta)\) to each line of the pair, then

\[ P_1P_2 = \frac{|a\alpha^2 + 2h\alpha\beta + b\beta^2|}{\sqrt{(a-b)^2 + 4h^2}} \]

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Proof:

Let \(y = m_1 x\) and \(y = m_2 x\) be lines represented by \(ax^2 + 2hxy + by^2 = 0\). Then

\[ m_1 + m_2 = \frac{-2h}{b} \quad \text{and} \quad m_1 m_2 = \frac{a}{b} \]

The distance of \((\alpha, \beta)\) from \(y = m_1 x\),

\[ P_1 = \frac{|m_1\alpha - \beta|}{\sqrt{1 + m_1^2}} \]

The distance of \((\alpha, \beta)\) from \(y = m_2 x\),

\[ P_2 = \frac{|m_2\alpha - \beta|}{\sqrt{1 + m_2^2}} \]

Thus,

\[\begin{align*} P_1P_2 &= \frac{|m_1\alpha - \beta|}{\sqrt{1 + m_1^2}} \cdot \frac{|m_2\alpha - \beta|}{\sqrt{1 + m_2^2}} \\ &= \frac{|m_1m_2\alpha^2 - (m_1 + m_2)\alpha\beta + \beta^2|}{\sqrt{(1 + m_1^2)(1 + m_2^2)}} \\ &= \frac{|m_1m_2\alpha^2 - (m_1 + m_2)\alpha\beta + \beta^2|}{\sqrt{1 + (m_1 + m_2)^2 + m_1^2 m_2^2}} \\ &= \frac{|m_1m_2\alpha^2 - (m_1 + m_2)\alpha\beta + \beta^2|}{\sqrt{1 + \left(\frac{-2h}{b}\right)^2 + \left(\frac{a}{b}\right)^2}} \\ &= \frac{\left|\frac{a}{b}\alpha^2 + \frac{2h}{b}\alpha\beta + \frac{b}{b}\beta^2\right|}{\sqrt{1 + \frac{4h^2}{b^2} - \frac{2a}{b} + \frac{a^2}{b^2}}} \\ &= \frac{|a\alpha^2 + 2h\alpha\beta + b\beta^2|}{\sqrt{b^2 + 4h^2 - 2ab + a^2}} \\ &= \frac{|a\alpha^2 + 2h\alpha\beta + b\beta^2|}{\sqrt{(a - b)^2 + 4h^2}} \end{align*}\]

This concludes the proof.

Example

Given the pair of straight lines \( 2x^2 - 3xy - y^2 = 0 \), find the length of the perpendiculars from the point \( (1, -2) \) to each line of the pair.

Solution:

The general formula for the product of the lengths of perpendiculars from a point \((\alpha, \beta)\) to the pair of straight lines \( ax^2 + 2hxy + by^2 = 0 \) is given by:

\[ P_1P_2 = \frac{|a\alpha^2 + 2h\alpha\beta + b\beta^2|}{\sqrt{(a - b)^2 + 4h^2}} \]

For the given equation \( 2x^2 - 3xy - y^2 = 0 \), we have:

\[ a = 2, \; h = -\frac{3}{2}, \; b = -1 \]

The point given is \((\alpha, \beta) = (1, -2)\).

Substituting these values into the formula, we get:

\[ \begin{align*} P_1P_2 &= \frac{|2(1)^2 + 2(-\frac{3}{2})(1)(-2) + (-1)(-2)^2|}{\sqrt{(2 - (-1))^2 + 4(-\frac{3}{2})^2}} \\ &= \frac{|2 + 6 - 4|}{\sqrt{(2 + 1)^2 + 4\left(\frac{9}{4}\right)}} \\ &= \frac{|4|}{\sqrt{3^2 + 9}} \\ &= \frac{4}{\sqrt{9 + 9}} \\ &= \frac{4}{\sqrt{18}} \\ &= \frac{4}{3\sqrt{2}} \\ &= \frac{4\sqrt{2}}{6} \\ &= \frac{2\sqrt{2}}{3} \end{align*} \]

Thus, the product of the lengths of the perpendiculars from the point \((1, -2)\) to the lines \( 2x^2 - 3xy - y^2 = 0 \) is:

\[ \boxed{\frac{2\sqrt{2}}{3}} \]