find the equation of a straight line on which the perpendicular from the origin makes an angle of 30 degree - Maths - Straight Lines

Question

find the equation of a straight line on which the perpendicular
from the origin makes an angle of 30 degree with the x-axis and
which forms a triangle of are 50/root3 with the axes

Gursheen kaur. · Accepted Answer

AB is the given line and OL = p is the perpendicular drawn from the origin to the line.

OL makes an angle of 30⁰ with OX.

$∴ α = 30^{0}$

Equation of line AB is given by

$x \cos α + y \sin α = p \Rightarrow x \cos 30^{0} + y \sin 30^{0} = p \Rightarrow \frac{x}{2} + \frac{\sqrt{3} y}{2} = p \Rightarrow x + \sqrt{3} y = 2 p ........ (1)$

Now, in ΔOLA,

$\cos 30^{0} = \frac{O L}{O A} \Rightarrow O A = \frac{2 p}{\sqrt{3}}$

Now, in ΔOLB,

$\cos 60^{0} = \frac{O L}{O B} \Rightarrow O B = 2 p$

Given that

$a r e a = \frac{50}{\sqrt{3}}$

$\Rightarrow \frac{1}{2} \times \frac{2 p}{\sqrt{3}} \times 2 p = \frac{50}{\sqrt{3}} \Rightarrow p^{2} = 25 \Rightarrow p = \pm 5$

Put this value in equation (1).

∴Equation of line AB is given by

$x + \sqrt{3} y = \pm 10$

find the equation of a straight line on which the perpendicular from the origin makes an angle of 30 degree with the x-axis and which forms a triangle of are 50/root3 with the axes.