
If the lines $y = \left( {2 + \sqrt 3 } \right)x + 4$ and $y = kx + 6$ are inclined at an angle $60^{\circ}$ to each other. Then what is the value of $k$?
A. 1
B. 2
C. $ - 1$
D. $ - 2$
Answer
232.8k+ views
Hint: The given equations of the lines are in the point-slope form. Find the slopes of the lines. Substitute the values of slopes in the formula of the angles between the two lines and get the required value of $k$.
Formula Used:
The angle between the two lines with slope ${m_1}$ and ${m_2}$ is: $\tan\theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$
The point-slope form of the line is: $y - {y_1} = m\left( {x - {x_1}} \right)$ , where $m$ is slope and $\left( {{x_1},{y_1}} \right)$ is the point on the line.
Complete step by step solution:
Given:
The equations of the lines are $y = \left( {2 + \sqrt 3 } \right)x + 4$ and $y = kx + 6$.
The angle between the lines is $60^{\circ}$.
Since the given equations of the lines are in the slope-intercept form of a line.
So, the slope of the line $y = \left( {2 + \sqrt 3 } \right)x + 4$ is,
${m_1} = \left( {2 + \sqrt 3 } \right)$
And the slope of the line $y = kx + 6$ is,
${m_2} = k$
Now apply the formula of the angle between the two lines.
$\tan\left( {60^{\circ}} \right) = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$
Substitute the values of the slopes in the above equation.
$\tan\left( {60^{\circ}} \right) = \left| {\dfrac{{k - \left( {2 + \sqrt 3 } \right)}}{{1 + k\left( {2 + \sqrt 3 } \right)}}} \right|$
$ \Rightarrow \sqrt 3 = \left| {\dfrac{{k - 2 - \sqrt 3 }}{{1 + 2k + \sqrt 3 k}}} \right|$ $\because \tan\left( {60^{\circ}} \right) = \sqrt 3 $ ]
$ \Rightarrow \sqrt 3 \left( {1 + 2k + \sqrt 3 k} \right) = k - 2 - \sqrt 3 $
Simplify the above equation.
$\sqrt 3 + 2\sqrt 3 k + 3k = k - 2 - \sqrt 3 $
$ \Rightarrow 2\sqrt 3 k + 2k = - 2 - 2\sqrt 3 $
$ \Rightarrow k\left( {2\sqrt 3 + 2} \right) = - \left( {2\sqrt 3 + 2} \right)$
$ \Rightarrow k = \dfrac{{ - \left( {2\sqrt 3 + 2} \right)}}{{\left( {2\sqrt 3 + 2} \right)}}$
$ \Rightarrow k = - 1$
Option ‘C’ is correct
Note: There are two formulas that are used to calculate the angle between the two lines.
Formula 1: $\tan\theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$, where ${m_1}$ and ${m_2}$ are the slopes of the lines.
Formula 2: If ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ are the two equations of the lines, then
the angle between the two lines is: $\tan\theta = \dfrac{{{a_2}{b_1} - {a_1}{b_2}}}{{{a_1}{a_2} + {b_1}{b_2}}}$
Formula Used:
The angle between the two lines with slope ${m_1}$ and ${m_2}$ is: $\tan\theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$
The point-slope form of the line is: $y - {y_1} = m\left( {x - {x_1}} \right)$ , where $m$ is slope and $\left( {{x_1},{y_1}} \right)$ is the point on the line.
Complete step by step solution:
Given:
The equations of the lines are $y = \left( {2 + \sqrt 3 } \right)x + 4$ and $y = kx + 6$.
The angle between the lines is $60^{\circ}$.
Since the given equations of the lines are in the slope-intercept form of a line.
So, the slope of the line $y = \left( {2 + \sqrt 3 } \right)x + 4$ is,
${m_1} = \left( {2 + \sqrt 3 } \right)$
And the slope of the line $y = kx + 6$ is,
${m_2} = k$
Now apply the formula of the angle between the two lines.
$\tan\left( {60^{\circ}} \right) = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$
Substitute the values of the slopes in the above equation.
$\tan\left( {60^{\circ}} \right) = \left| {\dfrac{{k - \left( {2 + \sqrt 3 } \right)}}{{1 + k\left( {2 + \sqrt 3 } \right)}}} \right|$
$ \Rightarrow \sqrt 3 = \left| {\dfrac{{k - 2 - \sqrt 3 }}{{1 + 2k + \sqrt 3 k}}} \right|$ $\because \tan\left( {60^{\circ}} \right) = \sqrt 3 $ ]
$ \Rightarrow \sqrt 3 \left( {1 + 2k + \sqrt 3 k} \right) = k - 2 - \sqrt 3 $
Simplify the above equation.
$\sqrt 3 + 2\sqrt 3 k + 3k = k - 2 - \sqrt 3 $
$ \Rightarrow 2\sqrt 3 k + 2k = - 2 - 2\sqrt 3 $
$ \Rightarrow k\left( {2\sqrt 3 + 2} \right) = - \left( {2\sqrt 3 + 2} \right)$
$ \Rightarrow k = \dfrac{{ - \left( {2\sqrt 3 + 2} \right)}}{{\left( {2\sqrt 3 + 2} \right)}}$
$ \Rightarrow k = - 1$
Option ‘C’ is correct
Note: There are two formulas that are used to calculate the angle between the two lines.
Formula 1: $\tan\theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$, where ${m_1}$ and ${m_2}$ are the slopes of the lines.
Formula 2: If ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ are the two equations of the lines, then
the angle between the two lines is: $\tan\theta = \dfrac{{{a_2}{b_1} - {a_1}{b_2}}}{{{a_1}{a_2} + {b_1}{b_2}}}$
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