Question

The slope of lines which makes an angle $60^\circ$ with the line $y - 3x + 18 = 0$ is

• A. \( \frac{3\sqrt{3} - 3}{1 + \sqrt{3}} \)
• B. \( \frac{3 + \sqrt{3}}{1 + \sqrt{3}} \)
• C. \( \frac{3}{1 + \sqrt{3}} \)
• D. \( \frac{\sqrt{3} - 1}{3}, \frac{\sqrt{3} + 1}{3} \)

Hint:

When finding the slope of a line that makes a given angle with another, use the formula for the tangent of the angle between two lines. This helps you derive the relationship between the slopes.

Answer 1

The Correct Option is B

The given equation is \( y - 3x + 18 = 0 \), which represents a line. The slope of the given line can be determined by converting the equation to slope-intercept form: \[ y = 3x - 18 \] Thus, the slope of this line is \( m_1 = 3 \). To find the slope of the lines that make a \( 60^\circ \) angle with the given line, we use the formula for the angle between two lines. If the angle between two lines is \( \theta \), the relationship between the slopes \( m_1 \) and \( m_2 \) is given by: \[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \] For \( \theta = 60^\circ \), \( \tan(60^\circ) = \sqrt{3} \). Hence, we solve the equation: \[ \sqrt{3} = \left| \frac{3 - m_2}{1 + 3m_2} \right| \] Solving this equation gives the values of \( m_2 \), which correspond to the slopes of the required lines. The correct slopes are found to be \( \frac{3 + \sqrt{3}}{1 + \sqrt{3}} \), which is option (B).