First slide

Equation of line in Straight lines

Question

A straight line $L = 0$ passes through the point $(5, - 4)$ is inclined at an angle of $60 °$ to the line $\sqrt{3} x + y - 3 = 0$ ,then the equation of the line $L = 0$ is

Moderate

A

$y + 4 = 0$
B

$\sqrt{3} x - y - 5 \sqrt{3} - 4 = 0$
C

$y + 4 = 0, \sqrt{3} x - y - 5 \sqrt{3} - 4 = 0$
D

$y - 4 = 0, \sqrt{3} x + y - 5 \sqrt{3} - 4 = 0$

Solution

$The equation of any line which makes an angle α with the line a x + b y + c = 0 and passing$
$through the point (x_{1}, y_{1}) is y - y_{1} = \tan (θ \pm α) (x - x_{1}) here \tan θ = - \frac{a}{b}$
$Here \tan θ = - \frac{\sqrt{3}}{1} \Rightarrow θ = 120^{\circ} and α = 60^{\circ}$
$Equations of the required lines are y + 4 = \tan (120^{\circ} \pm 60^{\circ}) (x - 5)$
It implies that
$\begin{matrix} y + 4 = \tan (180 °) (x - 5) \\ y + 4 = 0 \end{matrix}$
Or
$\begin{matrix} y + 4 = \tan (60 °) (x - 5) \\ y + 4 = \sqrt{3} (x - 5) \end{matrix}$
$Therefore, the equations of the required lines are y + 4 = 0, \sqrt{3} x - y - 5 \sqrt{3} - 4 = 0$

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