Linear Functions – Angle Between Lines
If the acute angle \( \theta \) between two straight lines \( y = m_1 x + A \) and \( y = m_2 x + B \), then
$$\tan{\theta} = \left|\frac{m_1 – m_2}{1+m_1 \times m_2}\right|$$
Basic of Angle Between Lines
Find the acute angles between the lines \( y = -2x + 4 \) and \( y = 3x + 2 \), giving your answer to the nearest degree.
\( \begin{aligned} \displaystyle
m_1 &= -2 \text{ and } m_2 = 3 \\
\tan \theta &= \left| \frac{-2-3}{1+(-2) \times 3}\right| \\
&= 1 \\
\theta &= \tan^{-1}1 \\
\therefore \theta &= 45^{\circ} \\
\end{aligned} \\ \)
Gradients for Angle Between Lines
Find the acute angles between the lines \( 2y = x + 1 \) and \( 2x – 3y = 4 \), giving your answer to the nearest degree.
\( \begin{aligned} \displaystyle
2y &= x + 1 \\
y &= \frac{1}{2}x + \frac{1}{2} \\
m_1 &= \frac{1}{2} \\
2x – 3y &= 4 \\
3y &= 2x + 4 \\
y &= \frac{2}{3}x + \frac{3}{4} \\
m_2 &= \frac{2}{3} \\
\tan{\theta} &= \left|\frac{\dfrac{1}{2}-\dfrac{2}{3}}{1+\dfrac{1}{2} \times \dfrac{2}{3}}\right| \\
&= \frac{1}{8} \\
\theta &= \tan^{-1} \frac{1}{8} \\
\therefore \theta &= 7^{\circ} \\
\end{aligned} \\ \)
The Nearest Minute for Angle Between Lines
Find the acute angle between the lines \( 2x+y+1=0 \) and \( x+y+4=0 \), correcting to the nearest minute.
\( \begin{aligned} \displaystyle
2x+y+1 &= 0 \\
y &= -2x-1 \\
m_1 &= -2 \\
x+y+4 &= 0 \\
y &= -x-1 \\
m_2 &= -1 \\
\tan{\theta} &= \left|\frac{-2-(-1)}{1+(-2) \times (-1)}\right| \\
&= \frac{1}{3} \\
\theta &= \tan^{-1}\frac{1}{3} \\
\therefore \theta &= 18^{\circ}26^{\prime} \\
\end{aligned} \\ \)
Two Points for Angle Between Lines
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.
\( \begin{aligned} \displaystyle \require{color}
2x-5y+1 &= 0 \\
5y &= 2x+1 \\
y &= \frac{2}{5}x + \frac{1}{5} \\
m_1 &= \frac{2}{5} \\
m_2 &= \frac{3-2}{5-(-1)} \\
&= \frac{1}{6} \\
\tan{\theta} &= \left|\frac{\dfrac{2}{5} – \dfrac{1}{6}}{1+\dfrac{2}{5} \times \dfrac{1}{6}}\right| \\
&= \frac{7}{32} \\
\theta &= \tan^{-1}\frac{7}{32} \\
\therefore \theta &= 12^{\circ}20^{\prime} \\
\end{aligned} \\ \)
Angle Between Lines of 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.
\( \begin{aligned} \displaystyle
e^{x} &= e^{-x} &\color{green} \text{find the intersection} \\
e^{x} \times e^{x} &= e^{-x} \times e^{x} \\
e^{2x} &= 1 \\
2x &= 0 \\
x &= 0 \\
\frac{dy_1}{dx} &= e^{x} \\
m_1 &= \left.\frac{dy_1}{dx}\right|_{x=0} \\
&= e^{0} \\
&= 1 \\
\frac{dy_2}{dx} &= -e^{x} \\
m_2 &= \left.\frac{dy_2}{dx}\right|_{x=0} \\
&= -e^{0} \\
&= -1 \\
\tan{\theta} &= \left|\frac{1+1}{1-1}\right| \\
&= \left|\frac{1+1}{0}\right| \\
&= \text{undefined} \\
\therefore \theta &= 90^{\circ} \\
\end{aligned} \\ \)
Angle Between Lines of 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.
\( \begin{aligned} \displaystyle \require{color}
\sin{2x} &= \sin{x} \\
2\sin{x}\cos{x} &= \sin{x} \\
\sin{x}(2\cos{x}-1) &= 1 \\
\sin{x} &= 0 \text{ or } 2\cos{x} – 1 = 0 \\
\sin{x} &\ne 0 &\color{green} 0 \lt x \lt 180^{\circ} \\
\cos{x} &= \frac{1}{2} \\
x &= 60^{\circ} \\
\frac{dy_1}{dx} &= \cos{x} \\
\left.\frac{dy_1}{dx} \right|_{x = 60^{\circ}} &= \cos{60^{\circ}} &\color{green} \\
m_1 &= \frac{1}{2} \color{green} \cdots (1) \\
\frac{dy_2}{dx} &= 2\cos{2x} \\
\left.\frac{dy_2}{dx}\right|_{x=60^{\circ}} &= 2\cos{120^{\circ}} \\
&= -1 \\
m_2 &= -1 \color{green} \cdots (2) \\
\tan\theta &= \left|\frac{\dfrac{1}{2}-(-1)}{1+\dfrac{1}{2} \times (-1)}\right| &\color{green} \text{by (1) and (2)} \\
&= 3 \\
\theta &= \tan^{-1}{3} \\
&= 72^{\circ} \\
\end{aligned} \\ \)
Angle Between Lines for Finding Gradients
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 1 decimal place.
\( \begin{aligned} \displaystyle \require{color}
\left|\frac{2-m}{1+2m}\right| &= \tan{25^{\circ}} \\
\frac{2-m}{1+2m} = \tan{25^{\circ}} &\text{ or } \frac{2-m}{1+2m} = -\tan{25^{\circ}} \\
\frac{2-m}{1+2m} &= \tan{25^{\circ}} \\
2-m &= \tan{25^{\circ}} + 2m\tan{25^{\circ}} \\
-m – 2m\tan{25^{\circ}} &= \tan{25^{\circ}} – 2 \\
m + 2m\tan{25^{\circ}} &= 2 – \tan{25^{\circ}} \\
m(1 + 2\tan{25^{\circ}}) &= 2 – \tan{25^{\circ}} \\
m &= \frac{2 – \tan{25^{\circ}}}{1 + 2\tan{25^{\circ}}} \\
m &= 0.8 \color{green} \cdots (1) \\
\frac{2-m}{1+2m} &= -\tan{25^{\circ}} \\
2-m &= -\tan{25^{\circ}} – 2m\tan{25^{\circ}} \\
-m + 2m\tan{25^{\circ}} &= -\tan{25^{\circ}} – 2 \\
2m\tan{25^{\circ}} – m &= \tan{25^{\circ}} + 2 \\
m(2\tan{25^{\circ}} – 1) &= \tan{25^{\circ}} + 2 \\
m &= \frac{\tan{25^{\circ}} + 2}{2\tan{25^{\circ}} – 1} \\
m &= 36.6 \color{green} \cdots (2) \\
\therefore m &= 0.8 \text{ or } m = 36.6 &\color{green} \text{by (1) and (2)} \\
\end{aligned} \\ \)
