Question

find the acute angle between the lines 2y+x=1 and x+3y=6

Answer 1

Given the equations of the lines:

$\begin{gathered} 2y+x=1 \\ x+3y=6 \end{gathered}$

the given equations written in standard form, we will rewrite in in the slope-intercept form

So,

$\begin{gathered} y=-\frac{1}{2}x+\frac{1}{2} \\ y=-\frac{1}{3}x+\frac{6}{3} \end{gathered}$

So, the slopes of the lines are { -1/2, -1/3 }

The acute angle between the lines of slopes m1 and m2 is given by the formula:

$\text{tan}\theta=|\frac{m_{1_{}}-m_2}{1+m_1m_2}|$

Substitute with slopes of the lines

So,

$\tan \theta=|\frac{-\frac{1}{2}-(-\frac{1}{3})}{1+(-\frac{1}{2})(-\frac{1}{3})}|=|\frac{-\frac{1}{6}}{1+\frac{1}{6}}|=\frac{1}{7}$

so, the angle will be:

$\theta=\tan ^{-1}(\frac{1}{7})=8.13\degree$

So, the answer will be the acute angle = 8.13°