First slide

Equation of line in Straight lines

Question

$A straight line L through the point (3, - 2) is inclined at an angle 60^{\circ} to the line \sqrt{3} x + y = 1 . If L also intersects$
$the x -axis, then the equation of L is$

Moderate

A

$y + \sqrt{3} x + 2 - 3 \sqrt{3} = 0$
B

$y - \sqrt{3} x + 2 + 3 \sqrt{3} = 0$
C

$\sqrt{3} y - x + 3 + 2 \sqrt{3} = 0$
D

$\sqrt{3} y + x - 3 + 2 \sqrt{3} = 0$

Solution

Let the slope of the required line be m. Then
$|\frac{m + \sqrt{3}}{1 - \sqrt{3} m}| = \sqrt{3}$
$\begin{aligned} or m + \sqrt{3} = \pm (\sqrt{3} - 3 m) \\ or 4 m = 0 or m = 0 \\ and 2 m = 2 \sqrt{3} or m = \sqrt{3} \end{aligned}$
Therefore the equation is
$\begin{array}{l} y + 2 = \sqrt{3} (x - 3) (m \neq 0 as given that line cuts the x -axis) \\ or \sqrt{3} x - y - (2 + 3 \sqrt{3}) = 0 \end{array}$

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