Which of these lines is perpendicular to the line $$ y=3 x+2 $$ ? $$ \begin{array}{l}\text { A } y=\frac{1}{3} x+3 \ \text { B } y=\frac{1}{3} x+2 \ \text { C } y=-\frac{1}{3} x+4 \ \text { D } y=3 x+\frac{1}{2}\end{array} $$

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To determine which line is perpendicular to the line $$ y = 3x + 2 $$, we need to consider the concept of slopes in perpendicular lines.

1. **Identify the slope of the given line:**
   - The equation $$ y = 3x + 2 $$ is in the slope-intercept form $$ y = mx + b $$, where $$ m $$ is the slope.
   - **Slope of the given line ($$ m_1 $$)**: $$ 3 $$.

2. **Determine the slope of the perpendicular line:**
   - Perpendicular lines have slopes that are **negative reciprocals** of each other.
   - **Slope of the perpendicular line ($$ m_2 $$)**: $$ -\frac{1}{3} $$.

3. **Check the slopes of the given options:**
   - **A.** $$ y = \frac{1}{3}x + 3 $$ ➔ Slope: $$ \frac{1}{3} $$ (Not perpendicular).
   - **B.** $$ y = \frac{1}{3}x + 2 $$ ➔ Slope: $$ \frac{1}{3} $$ (Not perpendicular).
   - **C.** $$ y = -\frac{1}{3}x + 4 $$ ➔ Slope: $$ -\frac{1}{3} $$ (**Perpendicular**).
   - **D.** $$ y = 3x + \frac{1}{2} $$ ➔ Slope: $$ 3 $$ (Not perpendicular).

**Answer:**  
**C. $$ y = -\frac{1}{3}x + 4 $$** is perpendicular to the line $$ y = 3x + 2 $$.
The line $$ y = -\frac{1}{3}x + 4 $$ is perpendicular to $$ y = 3x + 2 $$.

Which of these lines is perpendicular to the line ?

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