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

Correct Answer is : 2=√3rcosθ+2rsinθ

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

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

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

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

EAMCET Engineering 2003 Solved Paper

Section: Mathematics

Question : 111 of 160

Marks: +1, -0

The line passing through (−1,‌

) and perpendicular to √3sin‌θ+2‌cos‌θ=‌

is :

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