Show that the equation of a line passing through the origin and making an angle θ with the line y = mx + c is

^y/_x = ±^{(m + tan θ)}/_{(1 - m tan θ)}

Let the equation of the line passing through the origin be y = m₁x.

If this line makes an angle of θ with line y = mx + c, then

tanθ = |^{m₁ - m}/_{1 + m1m}|

⇒ tanθ = |^{(y/x - m)}/_{(1 + (y/x)m)}|

⇒ tanθ = |^{(y - mx)}/_{(x + my)}|

⇒ tanθ = ^{(y/x - m)}/_{(1 + (y/x)m)} or tan θ = ^{(y/x - m)}/_{(1 + (y/x)m)}

Case I: tanθ = ^{(y/x - m)}/_{(1 + (y/x)m)}

tanθ(1 + (y/x)m) = y/x - m

m + m tanθ = y/x (1 - m tanθ)

⇒ ^y/_x = ^{(m + tan θ)}/_{(1 - m tan θ)}

Case II: tanθ = ^{(m - y/x)}/_{(1 + (y/x)m)}

tanθ(1 + (y/x)m) = m - y/x

⇒ y/x (1 + m tan θ) = m - tanθ

⇒ ^y/_x = ^{(m - tan θ)}/_{(1 + m tan θ)}

Therefore, the required line is given by

^y/_x = ^{(m ± tan θ)}/_{(1 ∓ m tan θ)}