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 $L$ is

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

(A) $y+\sqrt{3} x+2-3 \sqrt{3}=0$

(C) $\sqrt{3} y-x+3+2 \sqrt{3}=0$

(D) $\sqrt{3} y+x-3+2 \sqrt{3}=0$

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Straight Lines

A straight line

L

through the point

(3, - 2)

is inclined at an angle

6 0^{\circ}

to the line

3 x + y = 1

. If

L

also intersects the

x

-axis, then the equation of

L

Answer: (B)

The given line

3 x + y = 1

has slope

m_{1} = - 3

.

Let the slope of line

L

m

. The angle between the two lines is

6 0^{\circ}

, so

\frac{m - ( - 3 )}{1 + m ( - 3 )} = tan 6 0^{\circ} = 3

.

This gives

\frac{m + 3}{1 - 3 m} = 3

, so

m + 3 = \pm 3 (1 - 3 m)

.

Case 1:

m + 3 = 3 - 3 m

gives

4 m = 0

, so

m = 0

.

Case 2:

m + 3 = - 3 + 3 m

gives

2 m = 23

, so

m = 3

.

Since

L

intersects the

x

-axis, it is not horizontal, so

m = 3

.

The equation is

y - (- 2) = 3 (x - 3)

, which simplifies to

y - 3 x + 2 + 33 = 0

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Single correct option