Question: Find the angle between the given lines: $√3x -y +2= 0 and x -√3y +3 = 0$.

Solution:
Given,

Equation of line (i) = $√3x -y +2=0$

Slope of line (i)  $(m_1) = - \dfrac{ coefficient \; of \;x}{coefficient \; of \;y}$
$= - \dfrac{√3}{-1}$
$= √3$
$= tan 60°$

Equation of line (ii) = $x - √3y +3 = 0$

Slope of line (ii) $(m_2) = -\dfrac{coefficient \; of \;x}{coefficient \; of \;y}$
$= -\dfrac{1}{-√3}$
$= -\dfrac{1}{√3}$
$= tan 30°$

Now,

Angle between two lines is given by:

$tan \theta = \left ( \pm \dfrac{m_1 -m_2}{1 +m_1×m_2} \right ) $

$or, tan \theta = \left ( \pm \dfrac{tan60° -tan30°}{1+ tan60°×tan30°} \right )$

$or, tan \theta = \pm tan (60°-30°)$

$or, tan \theta = \pm tan 30°$


Taking positive sign,

$or, tan \theta = tan30°$

$\therefore, \theta = 30°$


Taking negative sign,

$or, tan \theta = - tan 30°$

$or, tan \theta = tan (180° -30°)$

$\therefore, \theta = 150°

Therefore, the angle between the given pair of lines is either 30° or 150°.


Related Notes and Solutions:

Here is the website link to all the important formulae of Coordinate Geometry of Class 10.

#SciPiPupil