Angle between Two Lines

In the diagram, l₁ gradient m₁ and l₂ gradient m₂ makes angles θ₁ and θ₂ respectively with x-axis. The acute angle between the lines is \(\scriptsize \propto \)

m₁ = tan θ₁ , m₂ = tan θ₂

The exterior angle of triangle, θ₂ = θ₁ + \(\scriptsize \propto \)

∴ \(\scriptsize \propto \) = θ₂ – θ₁  

Then tanθ = tan(θ₂ – θ₁)

= \( \frac{tan \theta_2 \; – \; tan \theta_1 }{ 1 \; + \; tan \theta_2 tan \theta_1 } \)

= \( \frac{m_2 \; – \; m_1 }{ 1 \; + \; m_2 m_1 } \)

:- \(\scriptsize tan \propto \; = \normalsize \frac{m_2 \; – \; m_1 }{ 1 \; + \; m_2 m_1 } \)

Example:

Find the acute angle between the lines x + 4y = 12 and y – 2x + 6 = 0

Solution

Comparing the equations with y = mx+c 

the equation x + 4y = 12

\( \scriptsize \therefore 4y = \; -x \; + \; 12 \)

:- \( \scriptsize y =\normalsize – \frac{1}{4}\scriptsize x + \; 3 \)

∴gradient m = \(– \frac{1}{4} \)  

In the equation

y – 2x + 6 = 0

y = 2x – 6

∴ m = 2

Using the equation

:- \(\scriptsize tan \propto \; = \normalsize \frac{m_2 \; – \; m_1 }{ 1 \; + \; m_2 m_1 } \)

:- \(\scriptsize tan \propto \; = \normalsize \frac{-\frac{1}{4} \; – \; \frac{2}{1}}{ 1 \; + \; (\; -\frac{1}{4} \; \times \; 2) } \)

:- \(\scriptsize tan \propto \; = \normalsize \frac{-\frac{9}{4} }{ 1 \; – \; \frac{1}{2}} \)

:- \(\scriptsize tan \propto \; = \normalsize \frac{-\frac{9}{4} }{ \frac{1}{2}} \)

:- \(\scriptsize tan \propto \; = \normalsize \; – \frac{9}{4} \; \times \; \frac{2}{1}\)

:- \(\scriptsize tan \propto \; = \normalsize \; – \frac{9}{2} \)

:- \(\scriptsize \propto \; = tan^{-1} \left (\normalsize \frac{9}{2} \right) \)

:- \(\scriptsize \propto \; = 77.47^{0} \)