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.
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