Straight Lines
4.0 Angle between two lines
4.0 Angle between two lines
The acute angle $\theta $ between the two straight lines having slopes ${m_1}$ and ${m_2}$ is given by $$\theta = {\tan ^{ - 1}}\left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$$
CASE I: Two lines with slope ${m_1}$ and ${m_2}$ are parallel, then $$\begin{equation} \begin{aligned} \theta = 0{\text{ or }}\pi \Leftrightarrow \tan \theta = 0 \\ \therefore {m_1} = {m_2} \\\end{aligned} \end{equation} $$
- Two lines $ax+by+c=0$ and $a'x+b'y+c'=0$ are parallel if $\frac{a}{{a'}} = \frac{b}{{b'}} \ne \frac{c}{{c'}}$.
- The distance between two parallel lines $ax + by + {c_1} = 0$ and $ax + by + {c_2} = 0$ is $\left| {\frac{{{c_1} - {c_2}}}{{\sqrt {{a^2} + {b^2}} }}} \right|$.
CASE II: Two lines with slope ${m_1}$ and ${m_2}$ are perpendicular, then $$\begin{equation} \begin{aligned} \theta = \frac{\pi }{2}{\text{ or }} - \frac{\pi }{2} \Leftrightarrow \cot \theta = 0 \\ \therefore {m_1} \times {m_2} = - 1 \\\end{aligned} \end{equation} $$
- Two lines $ax+by+c=0$ and $a'x+b'y+c'=0$ are perpendicular if $aa'+bb'=0$.
Question 4. Find the equation of straight line which passes through the origin and making an angle of ${60^ \circ }$ with the line $x + \sqrt 3 y + 3\sqrt 3 = 0$.
Solution: Given line is $x + \sqrt 3 y + 3\sqrt 3 = 0$
$\begin{equation} \begin{aligned} \Rightarrow y = (\frac{{ - 1}}{{\sqrt 3 }})x - 3 \\ \therefore {\text{ Slope of given line = }}\frac{{ - 1}}{{\sqrt 3 }} \\\end{aligned} \end{equation} $
Let the slope of required line be $m$ and the angle between them is given i.e., ${60^ \circ }$
$$\begin{equation} \begin{aligned} \tan {60^ \circ } = \left| {\frac{{m - (\frac{{ - 1}}{{\sqrt 3 }})}}{{1 + m(\frac{{ - 1}}{{\sqrt 3 }})}}} \right| \\ \sqrt 3 = \left| {\frac{{\sqrt 3 m + 1}}{{\sqrt 3 - m}}} \right| \\ \pm \sqrt 3 = \frac{{\sqrt 3 m + 1}}{{\sqrt 3 - m}} \\\end{aligned} \end{equation} $$
Take $\sqrt 3 = \frac{{\sqrt 3 m + 1}}{{\sqrt 3 - m}}$, we get $m = \frac{1}{{\sqrt 3 }}$
Using $y=mx+c$, the equation of required line is $$y = \frac{1}{{\sqrt 3 }}x + c$$
Since it passes through origin i.e., $(0,0)$, $c=0$
Therefore, the equation of required line is $$y = \frac{1}{{\sqrt 3 }}x $$
Now, Take $-\sqrt 3 = \frac{{\sqrt 3 m + 1}}{{\sqrt 3 - m}}$, we get $m = \infty $ i.e., slope not defined.
Therefore, the equation of required line is $$x=0$$
