The line passing through $\left( { - 1,\frac{{\rm{\pi }}}{2}} \right)$ and perpendicular to $\sqrt 3 \sin {\rm{\theta }} + 2\cos {\rm{\theta }} = \frac{4}{r}$ is

$2 = \sqrt 3 r\cos {\rm{\theta }} - 2r\sin {\rm{\theta }}$

$5 = - 2\sqrt 3 r\sin {\rm{\theta }} + 4r\cos {\rm{\theta }}$

$2 = \sqrt 3 r\cos {\rm{\theta }} + 2r\sin {\rm{\theta }}$

$5 = 2\sqrt 3 r\sin {\rm{\theta }} + 4r\cos {\rm{\theta }}$

JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

The line passing through

\left( { - 1,\frac{\mathrm{\pi }}{2}} \right)

and perpendicular to

\sqrt 3 \sin \mathrm{\theta } + 2\cos \mathrm{\theta } = \frac{4}{r}

View answer and explanation

Correct option: A

2 = \sqrt 3 r\cos \mathrm{\theta } - 2r\sin \mathrm{\theta }

Option A (

2 = \sqrt 3 r\cos \mathrm{\theta } - 2r\sin \mathrm{\theta }

) is correct because it represents the line perpendicular to the given line and passing through the specified point in polar coordinates.