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^o 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 $ax + by + c = 0$ and passing through the point $(x_{1}, y_{1})$ is $y - y_{1} = \tan (θ \pm α) (x - x_{1})$ here $tanθ = - \frac{a}{b}$
Here $\tan θ = - \frac{\sqrt{3}}{1} \Rightarrow θ = 120 °$ and $α = 60 °$
Equations of the required lines are $y + 4 = \tan (120 ° \pm 60 °) (x - 5)$
It implies that
$\begin{array}{l} y + 4 = \tan (180 °) (x - 5) \\ y + 4 = 0 \end{array}$
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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