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. \( \frac{3 - \sqrt{3}}{1 + 3\sqrt{3}}, \frac{3 + \sqrt{3}}{1 - 3\sqrt{3}} \)
• 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

To find the possible slopes of lines that make a 60° angle with the given line, we follow these steps:

1. Determine Original Slope:
Given line equation: $y - 3x + 18 = 0$
Convert to slope-intercept form:
$y = 3x - 18$ → Slope $m_1 = 3$

2. Angle Between Lines Formula:
For two lines with slopes $m_1$ and $m_2$ making angle θ:
$\tanθ = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|$
Given θ = 60° and $\tan60° = \sqrt{3}$:
$\sqrt{3} = \left|\frac{3 - m_2}{1 + 3m_2}\right|$

3. Solve for Possible Slopes:
This absolute value equation gives two cases:

Case 1:
$\sqrt{3} = \frac{3 - m_2}{1 + 3m_2}$
Solving yields:
$m_2 = \frac{3 - \sqrt{3}}{1 + 3\sqrt{3}}$

Case 2:
$\sqrt{3} = -\frac{3 - m_2}{1 + 3m_2}$
Solving yields:
$m_2 = \frac{3 + \sqrt{3}}{1 - 3\sqrt{3}}$

Final Answer:
The possible slopes are:
$m_2 = \frac{3 - \sqrt{3}}{1 + 3\sqrt{3}}$ and $\frac{3 + \sqrt{3}}{1 - 3\sqrt{3}}$

Answer 2

To find the slope of lines making an angle of \(60^\circ\) with the given line \(y - 3x + 18 = 0\), we start by identifying the slope \(m_1\) of the given line. By rewriting the equation in the form \(y = mx + c\), we get:

\(y = 3x - 18\)

Thus, the slope \(m_1\) of this line is 3.

We know the formula for the tangent of the angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\):

\(\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1m_2}\right|\)

Given \(\theta = 60^\circ\), we have \(\tan 60^\circ = \sqrt{3}\). Therefore:

\(\sqrt{3} = \left|\frac{m_2 - 3}{1 + 3m_2}\right|\)

This equation splits into two cases:

Case 1:

\(\frac{m_2 - 3}{1 + 3m_2} = \sqrt{3}\)

Cross-multiplying:

\(m_2 - 3 = \sqrt{3}(1 + 3m_2)\)

\(m_2 - 3 = \sqrt{3} + 3\sqrt{3}m_2\)

Rearranging gives:

\(m_2 - 3\sqrt{3}m_2 = 3 + \sqrt{3}\)

\(m_2(1 - 3\sqrt{3}) = 3 + \sqrt{3}\)

\(m_2 = \frac{3 + \sqrt{3}}{1 - 3\sqrt{3}}\)

Case 2:

\(\frac{m_2 - 3}{1 + 3m_2} = -\sqrt{3}\)

Cross-multiplying:

\(m_2 - 3 = -\sqrt{3}(1 + 3m_2)\)

\(m_2 - 3 = -\sqrt{3} - 3\sqrt{3}m_2\)

Rearranging gives:

\(m_2 + 3\sqrt{3}m_2 = -\sqrt{3} + 3\)

\(m_2(1 + 3\sqrt{3}) = 3 - \sqrt{3}\)

\(m_2 = \frac{3 - \sqrt{3}}{1 + 3\sqrt{3}}\)

Thus, the possible slopes of the lines that make an angle of \(60^\circ\) with the line \(y - 3x + 18 = 0\) are:

\(m_2 = \frac{3 - \sqrt{3}}{1 + 3\sqrt{3}}, \quad m_2 = \frac{3 + \sqrt{3}}{1 - 3\sqrt{3}}\)