# Angle between Lines

If the acute angle $\theta$ between two straight lines $y = m_1 x + a$ and $y = m_2 x + b$, then
$$\tan{\theta} = \left|\frac{m_1 – m_2}{1+m_1 \times m_2}\right|$$

## Example 1: Straight Lines in Standard Form

Find the acute angles between the lines $y = -2x + 4$ and $y = 3x + 2$, giving your answer to the nearest degree.

\begin{aligned} \displaystyle m_1 &= -2 \text{ and } m_2 = 3 \\ \tan \theta &= \left| \frac{-2-3}{1+(-2) \times 3}\right| \\ &= 1 \\ \theta &= \tan^{-1}1 \\ \therefore \theta &= 45^{\circ} \\ \end{aligned} \\

## Example 2: Straight Lines in General Form

Find the acute angles between the lines $2y = x + 1$ and $2x – 3y = 4$, giving your answer to the nearest degree.

\begin{aligned} \displaystyle 2y &= x + 1 \\ y &= \frac{1}{2}x + \frac{1}{2} \\ m_1 &= \frac{1}{2} \\ 2x – 3y &= 4 \\ 3y &= 2x + 4 \\ y &= \frac{2}{3}x + \frac{3}{4} \\ m_2 &= \frac{2}{3} \\ \tan{\theta} &= \left|\frac{\dfrac{1}{2}-\dfrac{2}{3}}{1+\dfrac{1}{2} \times \dfrac{2}{3}}\right| \\ &= \frac{1}{8} \\ \theta &= \tan^{-1} \frac{1}{8} \\ \therefore \theta &= 7^{\circ} \\ \end{aligned} \\

## Example 3: Angle between Lines to the Nearest Minute

Find the acute angle between the lines $2x+y+1=0$ and $x+y+4=0$, correcting to the nearest minute.

\begin{aligned} \displaystyle 2x+y+1 &= 0 \\ y &= -2x-1 \\ m_1 &= -2 \\ x+y+4 &= 0 \\ y &= -x-1 \\ m_2 &= -1 \\ \tan{\theta} &= \left|\frac{-2-(-1)}{1+(-2) \times (-1)}\right| \\ &= \frac{1}{3} \\ \theta &= \tan^{-1}\frac{1}{3} \\ \therefore \theta &= 18^{\circ}26^{\prime} \\ \end{aligned} \\

## Example 4: Two Points and a Straight Line

Find the acute angle between the line $2x-5y+1=0$ and the line joining $(-1,2)$ and $(5,3)$, correcting to the nearest minute.

\begin{aligned} \displaystyle \require{color} 2x-5y+1 &= 0 \\ 5y &= 2x+1 \\ y &= \frac{2}{5}x + \frac{1}{5} \\ m_1 &= \frac{2}{5} \\ m_2 &= \frac{3-2}{5-(-1)} \\ &= \frac{1}{6} \\ \tan{\theta} &= \left|\frac{\dfrac{2}{5} – \dfrac{1}{6}}{1+\dfrac{2}{5} \times \dfrac{1}{6}}\right| \\ &= \frac{7}{32} \\ \theta &= \tan^{-1}\frac{7}{32} \\ \therefore \theta &= 12^{\circ}20^{\prime} \\ \end{aligned} \\

## Example 5: Angles at Intersection of Exponential Graphs

Find the angle between the tangents drawn to the curve $y_1 = e^{x}$ and $y_2 = e^{-x}$ at their point of intersection.

\begin{aligned} \displaystyle e^{x} &= e^{-x} &\color{red} \text{find the intersection} \\ e^{x} \times e^{x} &= e^{-x} \times e^{x} \\ e^{2x} &= 1 \\ 2x &= 0 \\ x &= 0 \\ \frac{dy_1}{dx} &= e^{x} \\ m_1 &= \left.\frac{dy_1}{dx}\right|_{x=0} \\ &= e^{0} \\ &= 1 \\ \frac{dy_2}{dx} &= -e^{x} \\ m_2 &= \left.\frac{dy_2}{dx}\right|_{x=0} \\ &= -e^{0} \\ &= -1 \\ \tan{\theta} &= \left|\frac{1+1}{1-1}\right| \\ &= \left|\frac{1+1}{0}\right| \\ &= \text{undefined} \\ \therefore \theta &= 90^{\circ} \\ \end{aligned} \\

## Example 6: Angles at Intersection of Trigonometric Graphs

Find the acute angle between the curves $y_1 = \sin{x}$ and $y_2 = \sin{2x}$, at their point of intersection, where $0 \lt x \lt 180^{\circ}$, correcting to the neatest degree.

\begin{aligned} \displaystyle \require{color} \sin{2x} &= \sin{x} \\ 2\sin{x}\cos{x} &= \sin{x} \\ \sin{x}(2\cos{x}-1) &= 1 \\ \sin{x} &= 0 \text{ or } 2\cos{x} – 1 = 0 \\ \sin{x} &\ne 0 &\color{red} 0 \lt x \lt 180^{\circ} \\ \cos{x} &= \frac{1}{2} \\ x &= 60^{\circ} \\ \frac{dy_1}{dx} &= \cos{x} \\ \left.\frac{dy_1}{dx} \right|_{x = 60^{\circ}} &= \cos{60^{\circ}} &\color{red} \\ m_1 &= \frac{1}{2} \color{red} \cdots (1) \\ \frac{dy_2}{dx} &= 2\cos{2x} \\ \left.\frac{dy_2}{dx}\right|_{x=60^{\circ}} &= 2\cos{120^{\circ}} \\ &= -1 \\ m_2 &= -1 \color{red} \cdots (2) \\ \tan\theta &= \left|\frac{\dfrac{1}{2}-(-1)}{1+\dfrac{1}{2} \times (-1)}\right| &\color{red} \text{by (1) and (2)} \\ &= 3 \\ \theta &= \tan^{-1}{3} \\ &= 72^{\circ} \\ \end{aligned} \\

## Example 7: Finding Gradients using Angle between Straight Lines

The acute angle between the lines $2x-y-7=0$ and $y=mx+3$ is $25^{\circ}$, find the value(s) of (m), correct to one decimal place.

\begin{aligned} \displaystyle \require{color} \left|\frac{2-m}{1+2m}\right| &= \tan{25^{\circ}} \\ \frac{2-m}{1+2m} = \tan{25^{\circ}} &\text{ or } \frac{2-m}{1+2m} = -\tan{25^{\circ}} \\ \frac{2-m}{1+2m} &= \tan{25^{\circ}} \\ 2-m &= \tan{25^{\circ}} + 2m\tan{25^{\circ}} \\ -m – 2m\tan{25^{\circ}} &= \tan{25^{\circ}} – 2 \\ m + 2m\tan{25^{\circ}} &= 2 – \tan{25^{\circ}} \\ m(1 + 2\tan{25^{\circ}}) &= 2 – \tan{25^{\circ}} \\ m &= \frac{2 – \tan{25^{\circ}}}{1 + 2\tan{25^{\circ}}} \\ m &= 0.8 \color{red} \cdots (1) \\ \frac{2-m}{1+2m} &= -\tan{25^{\circ}} \\ 2-m &= -\tan{25^{\circ}} – 2m\tan{25^{\circ}} \\ -m + 2m\tan{25^{\circ}} &= -\tan{25^{\circ}} – 2 \\ 2m\tan{25^{\circ}} – m &= \tan{25^{\circ}} + 2 \\ m(2\tan{25^{\circ}} – 1) &= \tan{25^{\circ}} + 2 \\ m &= \frac{\tan{25^{\circ}} + 2}{2\tan{25^{\circ}} – 1} \\ m &= 36.6 \color{red} \cdots (2) \\ \therefore m &= 0.8 \text{ or } m = 36.6 &\color{red} \text{by (1) and (2)} \\ \end{aligned} \\