Angle Between Two Lines - Practice Questions & MCQ

Angle between two straight line

\begin{aligned}
&\text { Two lines are given with the slope } m_1 \text { and } m_2 \text {, then acute angle } \theta \text { between the lines is given by }\\
&\theta=\tan ^{-1}\left|\frac{\mathrm{~m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|
\end{aligned}.

Let $\mathrm{m}_1$ and $\mathrm{m}_2$ be the slope of two given straight lines and $\theta_1$ and $\theta_2$ is the inclinations

$
\therefore \mathrm{m}_1=\tan \theta_1 \text { and } \mathrm{m}_2=\tan \theta_2
$

let $\theta$ and $\pi-\theta$ be the angles between straight line $\left(\theta \neq \frac{\pi}{2}\right)$
from the figure

$
\begin{aligned}
& \theta_2=\theta+\theta_1 \quad \text { or } \quad \theta=\theta_1-\theta_2 \\
& \tan (\theta)=\tan \left(\theta_1-\theta_2\right) \\
& \tan (\theta)=\left(\frac{\tan \left(\theta_2\right)-\tan \theta_1}{1+\tan \left(\theta_1\right) \tan \left(\theta_2\right)}\right)=\left(\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_2 \mathrm{~m}_1}\right)
\end{aligned}
$
Also, $\tan (\pi-\theta)=-\tan (\theta)=-\left(\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_2 \mathrm{~m}_1}\right)$

$
\Rightarrow \theta=\tan ^{-1}\left[ \pm\left(\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_2 \mathrm{~m}_1}\right)\right]
$
Hence, the acute angle between two straight lines is given as

$
\theta=\tan ^{-\mathbf{1}}\left|\left(\frac{\mathbf{m}_{\mathbf{2}}-\mathbf{m}_{\mathbf{1}}}{\mathbf{1}+\mathbf{m}_{\mathbf{2}} \mathbf{m}_{\mathbf{1}}}\right)\right|
$

Note:

1. If the angle between the two lines is $0^{\circ}$ or $\pi$ then lines are parallel two each other. In this case, $m_1=m_2$ where $m_1$ and $m_2$ are slopes of two lines.
2. If the angle between the two lines is $\frac{\pi}{2}$ or $-\frac{\pi}{2}$ then lines are perpendicular two each other. Then in this case $m_1 \cdot m_2=-1$ where, $m_1$ and $m_2$ are slopes of two lines.
3. Equation of two straight line given as $A_1 x+B_1 y+C_1=0$ and $A_2 x+B_2 y+C_2=0$. If these two lines are coincident then,

$
\frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{\mathrm{C}_1}{\mathrm{C}_2}
$
Illustriation
Find the angle between the line joining the points $(0,0),(2,6)$ with line joining the points $(2,3),(3,4)$
Let $A=(0,0), B=(2,6), C=(2,3)$ and $D=(3,4)$
Let $m_1$ is the slope of $A B$ and $m_2$ is slope of $C D$

$
\mathrm{m}_1=\frac{6-0}{2-0}=3 \text { and } \mathrm{m}_2=\frac{4-3}{3-2}=1
$
Let $\theta$ be the acute angle between the lines $A B$ and $C D$

$
\begin{aligned}
& \tan \theta=\left|\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|=\left|\frac{3-1}{1+3 \times 1}\right|=\frac{1}{2} \\
& \theta=\tan ^{-1}\left(\frac{1}{2}\right)
\end{aligned}
$