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Given two lines having slope $m_1$ and $m_2$, the angle between them is given by $\tan(\theta)=\frac{m_2- m_1}{1+m_1m_2}$

Does the order of $m_1$ and $m_2$ matter here, and if so what is the significance? Graphical aid would be helpful.

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  • $\begingroup$ What do you mean with order? $\endgroup$ Nov 4, 2016 at 20:54
  • $\begingroup$ Like would changing $m_2$-$m_1$ to $m_1$-$m_2$ change anything? I know it does....but I'm not sure what exactly.... $\endgroup$ Nov 4, 2016 at 20:56
  • $\begingroup$ It does: it changes the sign of $\tan(\theta)$; you can check that directly in the formula. This means there is an implicit positive orientation (counterclockwise) and a negative orientation (clockwise). When you reverse the order, you reverse orientation. $\endgroup$ Nov 4, 2016 at 21:02

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Let's see what happens if we switch $m_1$ and $m_2$. First call $\theta_1$ an angle such that $\tan\theta_1 = \frac{m_2-m_1}{1+m_1m_2}$ and $\theta_2$ such that $\tan\theta_2 = \frac{m_1-m_2}{1+m_2m_1}$. Immediately you can see that $\tan\theta_1 = -\tan\theta_2$ and by tangens being odd function, we get $\tan\theta_1 = \tan(-\theta_2)$. If we restrict our $\theta$'s to $\langle-\pi/2,\pi/2\rangle$ (which we usually do in this case since we want the acute angle between lines) we get that $\theta_2 = -\theta_1$ so it is "the same" angle but measured in opposite directions (clockwise or counterclockwise).

In conclusion, $\theta_1$ is the angle between lines $l_1$ and $l_2$ measured from line $l_1$ to $l_2$ and $\theta_2$ is the angle between $l_1$ and $l_2$ measured from $l_2$ to $l_1$. They are of the same absolute value, but opposite orientations.

If you don't want to be bothered with orientation, you can change formula to $\tan\theta =\left|\frac{m_2-m_1}{1+m_1m_2}\right|$ and you will always get a positive angle and switching $m_1$ and $m_2$ won't change it since $|m_2-m_1|=|m_1-m_2|$.

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  • $\begingroup$ @SaitamaSensei I think this is a correct, well explained answer to your question. If you are satisfied then the proper response in this site is to accept it (the check mark) and upvote it (the up arrow). If you are still confused you should try to explain why and get more help.The same comment applies to the other questions you've asked here. $\endgroup$ Jan 17, 2017 at 14:01
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Note that $m_1=\tan \theta_1$ and $m_1=\tan \theta_2$ are the tangents of the angles $\theta_1$ and $\theta_2$ between the lines and the $x$ axis. So: $$ \frac{m_2-m_1}{1+m_1m_2}=\frac{\tan \theta_2-\tan \theta_1}{1+\tan \theta_1\tan \theta_2}=\tan (\theta_2-\theta_1) $$ and: $$ \frac{m_1-m_2}{1+m_2m_1}=\frac{\tan \theta_1-\tan \theta_2}{1+\tan \theta_2\tan \theta_1}=\tan (\theta_1-\theta_2) $$ so the two formulas give two possible orientation for the angle $\theta$ between two lines.

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enter image description here

Consider two non-perpendicular and non-vertical intersecting lines $\overleftrightarrow{l_1}$ and $\overleftrightarrow{l_2}$ in the coordinate plane.

Observe that $\overleftrightarrow{l_1}$ and $\overleftrightarrow{l_2}$ form two pairs of vertically opposite angles (acute angles $a$ and $c$, obtuse angles $b$ and $d$) between them.

To express $a$ and $b$ in terms of the slopes $m_1$ and $m_2$, let’s first see how they are related to the inclinations of $\overleftrightarrow{l_1}$ and $\overleftrightarrow{l_2}$, i.e. $p$ and $q$. enter image description here

$$a=q-p$$ $$b= 180°-(q-p)$$

Now, since $p$ and $q$ are the inclinations of $\overleftrightarrow{l_1}$ and $\overleftrightarrow{l_2}$ respectively, their respective slopes are

$$m_1=\tan p$$ $$m_2=\tan q$$

So, to express a and b in terms of $m_1$ and $m_2$, let’s take the tan on both sides of (1) and (2) as follows: $$\tan a=\tan (q−p) and \tan b=\tan (180°−(q−p))$$ $$⇒ \tan a=\tan (q−p) and \tan b=−\tan (q−p)$$ [Using the result $\tan (180°−θ)=−\tan θ$.]

Using the above equations, we can express $\tan a and \tan b$ in terms of $m_1 and m_2$ as follows: $$ \begin{array}{l} \tan a=\dfrac{m_{2}-m_{1}}{1+m_{1} m_{2}} \\ \tan b=-\dfrac{m_{2}-m_{1}}{1+m_{1} m_{2}} \end{array} $$

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