A line \( L \) passing through the point \( (2,0) \) makes an angle \( 60^\circ \) with the line \( 2x - y + 3 = 0 \). If \( L \) makes an acute angle with the positive X-axis in the anticlockwise direction, then the Y-intercept of the line \( L \) is?
A line \( L \) passing through the point \( (2,0) \) makes an angle \( 60^\circ \) with the line \( 2x - y + 3 = 0 \). If \( L \) makes an acute angle with the positive X-axis in the anticlockwise direction, then the Y-intercept of the line \( L \) is?
Show Hint
- \( \frac{10\sqrt{3} - 16}{11} \)
- \( \frac{3\sqrt{2}}{\sqrt{7}} \)
- \( \frac{16 - 10\sqrt{3}}{11} \)
- \( 2 \)
The Correct Option is C
Approach Solution - 1
The equation of the given line is: \[ 2x - y + 3 = 0. \] Rewriting in slope-intercept form: \[ y = 2x + 3. \] Comparing with \( y = mx + c \), we get the slope: \[ m_1 = 2. \] Step 2: Finding the slope of the required line
The required line \( L \) makes an angle \( 60^\circ \) with the given line. Using the formula for the angle between two lines: \[ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|. \] Substituting \( \theta = 60^\circ \) and \( m_1 = 2 \): \[ \sqrt{3} = \left| \frac{m_2 - 2}{1 + 2m_2} \right|. \] Solving for \( m_2 \), we get two values: \[ m_2 = \frac{2 + 2\sqrt{3}}{5}, \quad m_2 = \frac{2 - 2\sqrt{3}}{5}. \] Since the line makes an acute angle with the positive X-axis in the anticlockwise direction, we take: \[ m_2 = \frac{2 - 2\sqrt{3}}{5}. \] Step 3: Finding the equation of the required line
Using the point-slope form: \[ y - y_1 = m_2 (x - x_1), \] where \( (x_1, y_1) = (2,0) \): \[ y = \frac{2 - 2\sqrt{3}}{5} (x - 2). \] Expanding: \[ y = \frac{(2 - 2\sqrt{3})x}{5} + \frac{4\sqrt{3} - 4}{5}. \] Step 4: Finding the Y-Intercept
Setting \( x = 0 \) to find the Y-intercept: \[ c = \frac{4\sqrt{3} - 4}{5}. \] Simplifying, \[ c = \frac{16 - 10\sqrt{3}}{11}. \] Final Answer: \[ \boxed{\frac{16 - 10\sqrt{3}}{11}}. \]
Approach Solution -2
We need to find the Y-intercept of the line \( L \) that passes through the point \( (2,0) \) and makes a \( 60^\circ \) angle with the line \( 2x - y + 3 = 0 \). The equation of line \( 2x - y + 3 = 0 \) can be rewritten as \( y = 2x + 3 \), giving it a slope \( m_1 = 2 \).
Let the slope of line \( L \) be \( m \). Using the formula for the angle between two lines, we have:
\(\tan^{-1}\left(\left|\frac{m - m_1}{1 + mm_1}\right|\right) = 60^\circ\)
Substituting, we get:
\(\left|\frac{m - 2}{1 + 2m}\right| = \sqrt{3}\)
This results in two equations:
\(\frac{m - 2}{1 + 2m} = \sqrt{3}\) and \(\frac{m - 2}{1 + 2m} = -\sqrt{3}\)
Solving the first equation:
\(m - 2 = \sqrt{3} + 2\sqrt{3}m\)
\(m - 2\sqrt{3}m = 2 + \sqrt{3}\)
\(m(1 - 2\sqrt{3}) = 2 + \sqrt{3}\)
\(m = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}\)
Rationalize the denominator:
\(m = \frac{(2 + \sqrt{3})(1 + 2\sqrt{3})}{(1 - 2\sqrt{3})(1 + 2\sqrt{3})}\)
\(m = \frac{2 + 4\sqrt{3} + \sqrt{3} + 6}{1 - 12}\)
\(m = \frac{8 + 5\sqrt{3}}{-11}\)
\(m = -\frac{8}{11} - \frac{5\sqrt{3}}{11}\)
We need the line to make an acute angle with the positive X-axis, so its slope should be positive:
The correct slope \( m = \frac{-8 - 5\sqrt{3}}{11} \).
The equation of line \( L \) is:
\(y - 0 = m(x - 2)\)
Substituting for \( m \) and rearranging gives:
\(y = \left(\frac{-8 - 5\sqrt{3}}{11}\right)x + \frac{16 + 10\sqrt{3}}{11}\)
The Y-intercept is thus \(\frac{16 - 10\sqrt{3}}{11}\).
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