Line l in the xy-plane contains points from each of Quadrants II, III, and IV, but no points from Quadrant I. Which of the following must be true?

Options

A. The slope of line l is undefined.
B. The slope of line l is zero.
C. The slope of line l is positive.
D. The slope of line l is negative.

Explanation

To solve the problem, we need to analyze the characteristics of line \( l \) in the xy-plane that passes through Quadrants II, III, and IV but does not intersect Quadrant I.

### Step-by-Step Solution:

1. **Understanding the Quadrants**:
- The xy-plane is divided into four quadrants:
- Quadrant I: \( (x > 0, y > 0) \)
- Quadrant II: \( (x < 0, y > 0) \)
- Quadrant III: \( (x < 0, y < 0) \)
- Quadrant IV: \( (x > 0, y < 0) \)

2. **Identifying the Line's Position**:
- Since the line contains points from Quadrants II, III, and IV, it must extend from the left side of the y-axis (Quadrant II) to the right side of the x-axis (Quadrant IV).
- The line cannot enter Quadrant I, which means it must either be horizontal or have a negative slope.

3. **Slope Calculation**:
- The slope \( m \) of a line is defined as the change in \( y \) divided by the change in \( x \) (i.e., \( m = \frac{\Delta y}{\Delta x} \)).
- To illustrate this, we can take two points on the line:
- Point A in Quadrant II: \( (-2, 4) \)
- Point B in Quadrant IV: \( (2, -4) \)

4. **Calculating the Slope**:
- Using the two points, we calculate the slope:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 4}{2 - (-2)} = \frac{-8}{4} = -2 \]
- The slope is negative, which confirms that the line is decreasing as it moves from Quadrant II to Quadrant IV.

5. **Conclusion**:
- Since the line has a negative slope and passes through Quadrants II, III, and IV but not Quadrant I, we can conclude that the statement about the slope being negative must be true.

### Final Answer:
The slope of the line \( l \) is negative.