Question

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 equal of L is:
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) $\sqrt 3 y + x - 3 + 2\sqrt 3 = 0$

Answer 1

Hint: According to given in the question we have to determine straight line L when a straight line L through the point $(3, - 2)$ is inclined at an angle ${60^\circ}$ to the line $\sqrt 3 x + y = 1$ and if L also intersects the x-axis. So, first of all we have to let the slope of the line L.
Now, we have to determine the slope of the given line $\sqrt 3 x + y = 1$ and as mentioned in the question that the inclined angle of line L is ${60^\circ}$ to the line $\sqrt 3 x + y = 1$. Hence, we have to find the slope of line L which we let.
Now, to find the slope of the lines $ax + by + c = 0$ we can use the formula as mentioned below:

Formula used: $ \Rightarrow m = - \dfrac{b}{a}$………………………..(A)
$ \Rightarrow \tan \theta = \dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}..............(B)$
Where, $\theta $is the angle at which the line is inclined and ${m_1}$ and ${m_2}$ are the slopes of the lines.
Now, as the equation of line L passes through the point $(3, - 2)$. Hence, with the help of the slope obtained for line L we can obtain the equation of the line L.
If a line $ax + by + c = 0$ passes through a point (a, b) and the slope is m then the equation of that line will be $(x - a) = m(y - b)$
Where, a and b are the points and m is the slope of that line.
$ \Rightarrow \tan {60^\circ} = \sqrt 3 ....................(C)$

Complete step-by-step answer:
Step 1: First of all we have to let the slope of line L as mentioned in the solution hint. Hence,
Slope of line L is = ${m_1}$
Step 2: Now, we have to determine the slope of the given line which is $\sqrt 3 x + y = 1$with the help of the formula as mentioned in the solution hint. Hence,
$
   \Rightarrow {m_2} = - \dfrac{{\sqrt 3 }}{1} \\
   \Rightarrow {m_2} = - \sqrt 3 \\
 $
Step 3: Now, to find the slope we let ${m_1}$for the line L we have to use the formula (B) as mentioned in the solution hint. Hence, on substituting all the values in the formula (B),
$ \Rightarrow \tan {60^\circ} = \dfrac{{{m_1} - ( - \sqrt 3 )}}{{1 + {m_1}( - \sqrt 3 )}}$………………(1)
Step 4: Now, to obtain the value of slope ${m_1}$ in the expression (1) we have to use the formula (c) as mentioned in the solution hint.
$ \Rightarrow \pm \sqrt 3 = \dfrac{{{m_1} + \sqrt 3 }}{{1 - \sqrt 3 {m_1}}}$
On applying cross-multiplication in the expression as obtained just above, and here, we take $ + \sqrt 3 $ to solve the expression.
$
   \Rightarrow \sqrt 3 (1 - \sqrt 3 {m_1}) = {m_1} + \sqrt 3 \\
   \Rightarrow \sqrt 3 - 3{m_1} = {m_1} + \sqrt 3 \\
   \Rightarrow - 3{m_1} - {m_1} = \sqrt 3 - \sqrt 3 \\
   \Rightarrow {m_1} = 0
 $
Now, we take $ - \sqrt 3 $to solve the expression. Hence,
$
   \Rightarrow - \sqrt 3 (1 - \sqrt 3 {m_1}) = {m_1} + \sqrt 3 \\
   \Rightarrow - \sqrt 3 + 3{m_1} = {m_1} + \sqrt 3 \\
   \Rightarrow 3{m_1} - {m_1} = \sqrt 3 + \sqrt 3 \\
   \Rightarrow 2{m_1} = 2\sqrt 3 \\
   \Rightarrow {m_1} = \sqrt 3
 $
Step 5: Now, as we know that a straight line L through the point $(3, - 2)$is inclined at an angle ${60^\circ}$to the line$\sqrt 3 x + y = 1$. Hence, on substituting all the values as explained in the solution hint.
$
   \Rightarrow y + 2 = \sqrt 3 (x - 3) \\
   \Rightarrow y + 2 = \sqrt 3 x - 3\sqrt 3 \\
   \Rightarrow y - \sqrt 3 x + 3\sqrt 3 + 2 = 0
 $
Final solution: Hence, with the help of formula (A), (B), and (C) we have obtained the required equation of line L = $y - \sqrt 3 x + 2 + 3\sqrt 3 = 0$.

Therefore option (B) is correct.

Note: One of the most important properties of a straight line is in how much its angle is away from the x-axis or the horizontal line and this concept is reflected some time called the slope of a line.
If the inclined angle of the line is given which is inclined at then we can determine the slope for that line with the help of some easy calculations.