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