Deriving the Equation of Perpendicular Lines Through the Origin

In this article, we will derive the equation of a pair of lines that pass through the origin and are perpendicular to the lines represented by the equation (2x^2 - 7xy - 5y^2 0). We will explore the mathematical steps involved and provide a detailed explanation of the algebraic manipulations required to find the desired equation.

Factorizing the Given Equation

The given equation is (2x^2 - 7xy - 5y^2 0). To factorize this equation, we look for two binomials that multiply to give the original equation. The factorization is:

(2x^2 - 7xy - 5y^2 (x - y)(2x 5y) 0)

This factorization agrees with the equation provided in the problem statement:

(text{answer is } xy 5x^2 y 0 text{ or } 5x^2 - 7xy 2y^2 0)

Revisiting the Factorization

The factorization can be represented as:

(5y^2/x^2 - 7y/x^2 0)

Let (y/x m) be the slope of one of the lines. To find the line perpendicular to it, we need the slope (-1/m). Substituting (y/x -x/y) into the equation, we get:

(5(-x/y)^2 - 7(-x/y)^2 - 2 0)

This simplifies to:

(2y^2 - 7xy 5x^2 0)

Using the Auxiliary Equation

The auxiliary equation of the given pair of lines (L_1) and (L_2) is:

(5m^2 - 7m 2 0)

The slopes of the lines are the roots of this auxiliary equation:

(m - 1 0 Rightarrow m_1 1)

(5m - 2 0 Rightarrow m_2 frac{2}{5})

The slopes of the lines perpendicular to (L_1) and (L_2) are (-1/m_1 -1) and (-1/m_2 -frac{5}{2}). The equations of these lines are:

(y -x Rightarrow x y 0)

(y -frac{5}{2}x Rightarrow 5x 2y 0)

The joint equation of these lines is:

(x y - 5x - 2y 0 Rightarrow 5x^2 - 7xy 2y^2 0)

Using Rotation and Perpendicular Lines

To find the equation of the lines perpendicular to (2x^2 - 5xy - 2y^2 0), we perform a 90-degree rotation by substituting (x) with (-y) and (y) with (x). The equation becomes:

(2(-y)^2 - 5(-y)(x) - 2(x)^2 0 Rightarrow 2y^2 5xy - 2x^2 0)

Now, the perpendicular lines that pass through the origin are (x 2y 0) and (2x y 0). To find the joint equation, we multiply these two equations:

((x 2y)(2x y) 0 Rightarrow 2x^2 xy 4xy 2y^2 0 Rightarrow 2x^2 5xy 2y^2 0)

This matches the derived equation (5x^2 - 7xy 2y^2 0), confirming the result.

Conclusion

The single equation of the pair of lines passing through the origin and perpendicular to the lines represented by (2x^2 - 5xy - 2y^2 0) is:

(boxed{5x^2 - 7xy 2y^2 0})

This solution uses algebraic manipulations and the properties of perpendicular lines to find the desired equation.