JEE MATHEMATICS MCQ: STRAIGHT LINES

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Application +4 | -1

The equation of line passing through

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

and perpendicular to

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

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it represents the equation of a line passing through the given point and perpendicular to the given equation.

More Questions from STRAIGHT LINES

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x-3 y=p, a x+2 y=q

a x+y=r

\left\{ {a{t_1}{t_2},\;a\left( {{t_1} + {t_2})\} ,\{ a{t_2}{t_3},a({t_2} + {t_3}} \right)} \right\},\;\left\{ {a{t_3}{t_1},\;a\left( {{t_3} + {t_1}} \right)} \right\}

y = 2\;x + 4

6x = 3y + 5

{x^2} + {y^2} + 2xy - 8ax - 8ay - 9{a^2} = 0

(3,8)

x+3 y=7

\left( {{x^2} + 7xy + 2{y^2}} \right)\left( {y - 1} \right) = 0