JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Medium

1 of 1

Application +4 | -1

A line passing through the point

A(9,0)

makes angle of

30^{\circ}

with the positive direction of

x

-axis. If this line is rotated about

A

through an angle of

15^{\circ}

in the clockwise direction, then its equation in the new position is :

View answer and explanation

Correct option: A

\frac{y}{\sqrt{3}-2}+x=9

Option A (

\frac{y}{\sqrt{3}-2}+x=9

) is correct because after rotating the line by

15^{\circ}

clockwise, the slope becomes

\tan(15^{\circ}) = 2 - \sqrt{3}

, and the equation is derived using the point-slope form.

More Questions from STRAIGHT LINES

Practice All Chapter MCQs →

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