Browse Questions

# Find angles between the lines $\sqrt 3x+y=1$ and $x+\sqrt 3y=1$.

$\begin{array}{1 1}(A)\;30^{\circ} \: or \: 150^{\circ} \\(B)\; 45^{\circ} \: or \: 315^{\circ} \\(C)\; 60^{\circ} \: or \: 120^{\circ} \\(D)\;0^{\circ} \end{array}$

Toolbox:
• Angle between two lines is $\theta = \tan^{-1} \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$ where $m_1$ and $m_2$ are the slopes of the two lines.
The given lines are $\sqrt 3x+y=1$ -----------(1) and
$\qquad \qquad \qquad \qquad x+\sqrt 3y=1$----------(2)
Slope of the line (1) is $m_1=-\sqrt 3$ and
Slope of the line (2) is $m_2= -\large\frac{1}{\sqrt 3}$.
$\therefore \tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
$= \bigg| \large\frac{-\sqrt 3+\Large\frac{1}{\sqrt 3}}{1+(-\sqrt 3)\bigg( -\Large\frac{1}{\sqrt 3} \bigg) } \bigg|$
$= \bigg| \large\frac{\Large\frac{-3+1}{\sqrt 3}}{1+1} \bigg|$
$\therefore \tan \theta = \bigg| \large\frac{1}{\sqrt 3} \bigg|$
$\Rightarrow \theta = 30^{\circ}\: or \: 180^{\circ}-30^{\circ}=150^{\circ}$.
Hence the angle between the two lines is $30^{\circ}$