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.