A straight line L through the point (3,-2) is inclined at an angle 60^ to the line 3 x+y=1 . If L also intersects the X -axis, then the equation of L is

Correct answer: B. y- 3 x+2+3 3 =0 A straight line passing through P and making an angle of α=60^ , is given by (y-y_1)/(x-x_1)= (θ ± α) where, 3 x+y=1 y=- 3 x+1 Then, θ=- 3 array ll & (y+2)/(x-3)=( θ ± α)/(1 θ α) & (y+2)/(x-3)= - 3 + 3 1-(- 3 )( 3 ) and & (y+2)/(x-3)= - 3 - 3 1+(- 3 )( 3 ) array array ll & y+2=0 and & (y+2)/(x-3)= -2 3 1-3 = 3 & y+2= 3 x-3 3 array Neglecting, y+2=0 as it does not intersect Y -axis.

JEE Advanced 2011Paper 1

Exam: JEE Advanced
Year: 2011
Paper: Paper 1
Subject: Mathematics
Chapter: Straight Lines

JEE Advanced 2011 · Paper 1 · Mathematics · Straight Lines

A straight line L through the point (3,-2) is inclined at an angle...

A straight line

L

through the point

(3,-2)

is inclined at an angle

60^{\circ}

to the line

\sqrt{3} x+y=1

. If

L

also intersects the

X

-axis, then the equation of

L

Options

A $y+\sqrt{3} x+2-3 \sqrt{3}=0$
B $y-\sqrt{3} x+2+3 \sqrt{3}=0$
C $\sqrt{3} y-x+3+2 \sqrt{3}=0$
D $\sqrt{3} y+x-3+2 \sqrt{3}=0$

Correct answer

B. $y-\sqrt{3} x+2+3 \sqrt{3}=0$

Step-by-step solution

A straight line passing through

P

and making an angle of

\alpha=60^{\circ}

, is given by

\frac{y-y_1}{x-x_1}=\tan (\theta \pm \alpha)

where,

\quad \sqrt{3} x+y=1

\Rightarrow \quad y=-\sqrt{3} x+1 Then,

\quad \tan \theta=-\sqrt{3}

\begin{array}{ll} \Rightarrow & \frac{y+2}{x-3}=\frac{\tan \theta \pm \tan \alpha}{1 \mp \tan \theta \tan \alpha} \\ \Rightarrow & \frac{y+2}{x-3}=\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3})(\sqrt{3})} \\ \text{ and } & \frac{y+2}{x-3}=\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})} \end{array}

\begin{array}{ll} \Rightarrow & y+2=0 \\ \text{ and } & \frac{y+2}{x-3}=\frac{-2 \sqrt{3}}{1-3}=\sqrt{3} \\ \Rightarrow & y+2=\sqrt{3} x-3 \sqrt{3} \end{array}

Neglecting,

y+2=0

as it does not intersect

Y

-axis.

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