Question

Two first-degree equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 represent the same straight line if and only if a₁:b₁:c₁ = a₂:b₂:c₂.

Answer 1

Suppose the two equations represent the same straight line. Since every line is parallel to itself, a₁b₂ = a₂b₁ and hence a₁:b₁ = a₂:b₂. If b₁ = 0, then b₂ = 0 so that a₁ and a₂ are non-zero.

This implies that 

-c₁/a₁ = - c₂/a₂

 Therefore

 a₁: b₁: c₁ =  a₂ : b₂ : c₂

 If  b₁x ≠ 0, then b₂ ≠ x  0(since a₁b₂ = a₂b₁). Let

b₁/b₂ = λ

 Therefore,

a₁b₂ = a₂b₁ ⇒ a₂/a₁ = b₂/b₁ = 1/λ

⇒  a₁ = λa₂   ...(1)

 Also (0, −c₁/b₁) is a point on a₁x + b₁y = c₁ = 0 which implies that(0, -c₁ /b₁) also lies on a₂x + b₂y + c_{2 =} 0. 

Therefore

Therefore, from Eqs. (1) and (2)

 a₁: b₁: c₁ =  a₂ : b₂ : c₂

Conversely, suppose a₁: b₁: c₁ =  a₂ : b₂ : c₂ Therefore, for some real  λ ≠ 0, we have a₁ = λa₂, b₁ = λb₂, c₁ = λc₂. Hence

Therefore, both equations represent the same straight line.