Question

Find the acute angle between the pair of lines $x + 3y + 5 = 0$ and $2x + y - 1 = 0$ .

Answer 1

Simplify the above given equations in $y = mx + c$ format and find slopes of both lines. After that use the formula $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$ to get the angles between the lines where $m_1$ and $m_2$ are the slopes of lines.

Complete step-by-step answer:
The equation of first line is
$x + 3y + 5 = 0$

$$\therefore 3y = - x - 5 \\\ \therefore y = - \dfrac{x}{3} - \dfrac{5}{3} \\\$$

On comparing the above equation by $y = mx + c$ , we get the slope of the first line as ${m_1} = - \dfrac{1}{3}$ .
The equation of second line is
$2x + y - 1 = 0$
$\therefore y = - 2x + 1$
On comparing the above equation by $y = mx + c$ , we get the slope of the second line as ${m_2} = - 2$ .
Now, for angle between the lines, we use the formula $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$ .
$\therefore \tan \theta = \left| {\dfrac{{ - \dfrac{1}{3} - \left( { - 2} \right)}}{{1 + \left( { - \dfrac{1}{3}} \right)\left( { - 2} \right)}}} \right|$
$\therefore \tan \theta = \left| {\dfrac{{ - \dfrac{1}{3} + 2}}{{1 + \dfrac{2}{3}}}} \right|$
$\therefore \tan \theta = \left| {\dfrac{{\dfrac{{ - 1 + 6}}{3}}}{{\dfrac{{3 + 2}}{3}}}} \right|$

$$\therefore \tan \theta = \left| {\dfrac{5}{5}} \right| \\\ \therefore \tan \theta = \left| 1 \right| \\\ \therefore \theta = {\tan ^{ - 1}}\left( 1 \right) \\\ \therefore \theta = 45^\circ \\\$$

Thus, the angle between the lines is $45^\circ$ .

Note: The slope of a line can also be written as $m = \tan \theta$ .
So, here ${m_1} = \tan {\theta _1} = - \dfrac{1}{3}$ and ${m_2} = \tan {\theta _2} = - 2$ .
Instead of $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$ , you can also use the formula $\tan \left( {{\theta _1} - {\theta _2}} \right) = \left| {\dfrac{{\tan {\theta _1} - \tan {\theta _2}}}{{1 + \tan {\theta _1}\tan {\theta _2}}}} \right|$ to find the angle between the lines.