Question

Prove that homogeneous equation of degree two in x and y, ax² + 2hxy + by² = 0 represents a pair of lines passing through the origin if h² − ab ≥ 0. Hence show that equation x² + y² = 0 does not represent a pair of lines.

Answer 1

A homogeneous equation of degree two in x and y is:

ax² + 2hxy + by² = 0

This equation represents a pair of straight lines passing through the origin if it can be factored into two linear factors of the form:

(l₁​x + m₁​y) (l₂​x + m₂​y) = 0

Where l₁, m₁, l₂, m₂​ are constants.

To check whether the given equation represents a pair of lines, we treat it as a quadratic equation in `x/y` or `y/x` by assuming:

ax² + 2hxy + by² = 0

Dividing throughout by y² (assuming y ≠ 0):

`a(x/y)^2 + 2h(x/y) + b = 0`

This is a quadratic equation in `x/y`. For this to represent two distinct lines, it must have two distinct real roots. The condition for real roots is that the discriminant must be non-negative.

(2h)² − 4ab ≥ 0

4h² − 4ab ≥ 0

h² − ab ≥ 0

Thus, the equation ax² + 2hxy + by² = 0 represents a pair of straight lines through the origin if and only if:

h² − ab ≥ 0

Consider the equation:

x² + y² = 0

Comparing with the general form ax² + 2hxy + by² = 0:

we have a = 1, h = 0, b = 1

Now, check the condition:

h² − ab = 0² − (1 × 1) = 0 − 1 = −1

Since h² − ab < 0, the equation x² + y² = 0 does not satisfy the condition for representing a pair of lines. Instead, the only real solution to this equation is x = 0, y = 0, meaning it represents only the origin as a single point, not two lines.