https://file.aiquickdraw.com/mathbot/file/1724375356545sd7s27oq.jpg

Let's work through the exercises in the image one by one.

### **22. Marquen con una X las rectas paralelas a \( y = \frac{1}{2}x + 5 \):**

- **Concept**: Two lines are parallel if they have the same slope.

Given line: \( y = \frac{1}{2}x + 5 \). The slope is \( \frac{1}{2} \).

Now, let's check the slopes of the other lines:
- \( y = 2x + 3 \): Slope is 2 (not parallel).
- \( y = \frac{1}{2}x - 3 \): Slope is \( \frac{1}{2} \) (parallel).
- \( y = -2x + 1 \): Slope is -2 (not parallel).
- \( y = \frac{1}{2}x + 2 \): Slope is \( \frac{1}{2} \) (parallel).

**Answer**: The parallel lines are \( y = \frac{1}{2}x - 3 \) and \( y = \frac{1}{2}x + 2 \).

### **23. Marquen con una X las rectas perpendiculares a la recta \( y = -2x + 9 \):**

- **Concept**: Two lines are perpendicular if the product of their slopes is \( -1 \).

Given line: \( y = -2x + 9 \). The slope is \( -2 \).

To find perpendicular lines, their slopes must be the negative reciprocal of -2, which is \( \frac{1}{2} \).

Let's check the slopes of the other lines:
- \( y = -2x + 3 \): Slope is -2 (not perpendicular).
- \( y = \frac{1}{2}x - 3 \): Slope is \( \frac{1}{2} \) (perpendicular).
- \( y = 2x + 1 \): Slope is 2 (not perpendicular).
- \( y = -\frac{1}{2}x + 3 \): Slope is \( -\frac{1}{2} \) (not perpendicular).

**Answer**: The perpendicular line is \( y = \frac{1}{2}x - 3 \).

### **24. Completen con \( || \) o \( \perp \), según corresponda:**

- \( R_1: y = 3x - 1 \) and \( R_2: y = -\frac{1}{3}x + 2 \)
  - Slopes: 3 and \( -\frac{1}{3} \). \( 3 \times -\frac{1}{3} = -1 \), so they are perpendicular ( \( \perp \) ).

- \( R_1: y = -\frac{1}{2}x + 3 \) and \( R_2: y = 2x - 1 \)
  - Slopes: \( -\frac{1}{2} \) and 2. \( -\frac{1}{2} \times 2 = -1 \), so they are perpendicular ( \( \perp \) ).

- \( R_1: y = -3x + 7 \) and \( R_2: y = -3x + 5 \)
  - Slopes: Both are -3, so they are parallel ( \( || \) ).

### **25. Escriban la ecuación de la recta que cumple con lo pedido en cada caso:**

- **a)** **Paralela a** \( y = 2x - 5 \) **y que pase por el punto** \( (-2, 1) \):
  - The slope should be the same as \( y = 2x - 5 \), which is 2.
  - Equation form: \( y - y_1 = m(x - x_1) \) where \( m = 2 \), \( x_1 = -2 \), and \( y_1 = 1 \).
  - \( y - 1 = 2(x + 2) \)
  - \( y - 1 = 2x + 4 \)
  - \( y = 2x + 5 \)

- **b)** **Perpendicular a** \( y = -\frac{1}{2}x + 3 \) **y que pase por el punto** \( (4, -3) \):
  - The slope should be the negative reciprocal of \( -\frac{1}{2} \), which is 2.
  - Equation form: \( y - y_1 = m(x - x_1) \) where \( m = 2 \), \( x_1 = 4 \), and \( y_1 = -3 \).
  - \( y + 3 = 2(x - 4) \)
  - \( y + 3 = 2x - 8 \)
  - \( y = 2x - 11 \)

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Would you like any additional details or explanations on any part? 

Here are some related questions for further exploration:

1. How do you determine the slope from the equation of a line?
2. What is the significance of parallel and perpendicular lines in geometry?
3. Can you explain why the slopes of perpendicular lines multiply to \(-1\)?
4. How would you find the equation of a line given two points?
5. How does the y-intercept affect the position of a line on a graph?

**Tip**: Remember that in any equation of the form \( y = mx + b \), \( m \) represents the slope, which is crucial for identifying relationships between lines.

Explore geometry problems involving parallel and perpendicular lines. Learn how to determine parallel and perpendicular lines based on their slopes and equations. Detailed solutions provided for each problem. Suitable for Grades 7-9 students.

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