Question

Show that the joint equation of a pair of straight lines through the origin is a homogeneous equation of second degree in x and y.

Answer 1

Let the two lines through the origin be a₁x + b₁y = 0 and a₂x + b₂y = 0.

Consider the locus represented by (a₁x + b₁y)(a₂x + b₂y) = 0   ...(1)

If (x₁, y₁) be any point on the line a₁x + b₁y = 0, then a₁x₁ + b₁y₁ = 0

∴ (a₁x₁ + b₁y₁)(a₂x₁ + b₂y₁) = 0

This shows that any point on the line a₁x + b₁y = 0 lies on the locus represented by (1).

Similarly, we can show that any point (x₂, y₂) on the line a₂x + b₂y = 0 also lies on the locus represented by (1).

So, all the points on the lines a₁x + b₁y = 0 and a₂x + b₂y = 0 lie on the locus (1),

i.e. locus (1) contains both these lines.  ... (A)

Conversely, if (x₀, y₀) is any point on the locus (1), then

(a₁x₀ + b₁y₀)(a₂x₀ + b₂y₀) = 0

∴ a₁x₀ + b₁y₀ = 0 or a₂x₀ + b₂y₀ = 0 or both are zero.

This shows that any point on the locus (1) lies on the line a₁x + b₁y = 0 or on the line a₂x + b₂y = 0 or on both the lines.

Hence, from (A) and (B), it follows that equation (1) represents both the lines a₁x + b₁y = 0 and a₂x + b₂y = 0 jointly, i.e. equation (1) is the joint equation of the pair of lines.

The equation (a₁x + b₁y)(a₂x + b₂y) = 0 can be written as:

a₁a₂x² + (a₁b₂ + a₂b₁)xy + b₁b₂y² = 0

Let a₁a₂ = a, a₁b₂ + a₂b₁ = 2h and b₁b₂ = b, then the joint equation is ax² + 2hxy + by² = 0 which is the homogeneous equation of second degree in x and y.

Hence, the joint equation of the pair of lines passing through the origin is a homogeneous equation of second degree in x and y.