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Let's break down and solve each part of the given problem.

### Problem: 
Find the equation for the line that:

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#### a) Is parallel to the line defined by \( y = 2x + 3 \) and passes through the point \( (4, 5) \).

1. **Parallel lines** have the same slope. The given line \( y = 2x + 3 \) is in slope-intercept form \( y = mx + b \), where the slope \( m = 2 \).
   
2. The equation of a line with slope \( m = 2 \) passing through \( (4, 5) \) can be written using the point-slope form:
   \[
   y - y_1 = m(x - x_1)
   \]
   Substituting \( m = 2 \), \( x_1 = 4 \), and \( y_1 = 5 \):
   \[
   y - 5 = 2(x - 4)
   \]
   Expanding:
   \[
   y - 5 = 2x - 8 \quad \Rightarrow \quad y = 2x - 3
   \]
   **Equation: \( y = 2x - 3 \)**

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#### b) Is parallel to the line defined by \( y = x + 1 \) and passes through the point \( (1, 3) \).

1. The given line has a slope of \( m = 1 \) (since \( y = x + 1 \) means \( m = 1 \)).

2. Using the point-slope form with slope \( m = 1 \) and point \( (1, 3) \):
   \[
   y - 3 = 1(x - 1)
   \]
   Expanding:
   \[
   y - 3 = x - 1 \quad \Rightarrow \quad y = x + 2
   \]
   **Equation: \( y = x + 2 \)**

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#### c) Is perpendicular to the line defined by \( y = -\frac{1}{2}x + 2 \) and passes through the point \( (3, -5) \).

1. **Perpendicular lines** have slopes that are negative reciprocals. The slope of the given line is \( m = -\frac{1}{2} \), so the slope of the perpendicular line will be \( m = 2 \) (negative reciprocal of \( -\frac{1}{2} \)).

2. Using the point-slope form with slope \( m = 2 \) and point \( (3, -5) \):
   \[
   y - (-5) = 2(x - 3) \quad \Rightarrow \quad y + 5 = 2(x - 3)
   \]
   Expanding:
   \[
   y + 5 = 2x - 6 \quad \Rightarrow \quad y = 2x - 11
   \]
   **Equation: \( y = 2x - 11 \)**

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#### d) Is perpendicular to the line defined by \( y = \frac{1}{2}x - 5 \) and passes through the point \( (-2, -6) \).

1. The slope of the given line is \( m = \frac{1}{2} \), so the slope of the perpendicular line will be \( m = -2 \) (negative reciprocal of \( \frac{1}{2} \)).

2. Using the point-slope form with slope \( m = -2 \) and point \( (-2, -6) \):
   \[
   y - (-6) = -2(x - (-2)) \quad \Rightarrow \quad y + 6 = -2(x + 2)
   \]
   Expanding:
   \[
   y + 6 = -2x - 4 \quad \Rightarrow \quad y = -2x - 10
   \]
   **Equation: \( y = -2x - 10 \)**

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### Recap of Equations:
- a) \( y = 2x - 3 \)
- b) \( y = x + 2 \)
- c) \( y = 2x - 11 \)
- d) \( y = -2x - 10 \)

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Would you like further details on any of these parts, or have any other questions?

### Related Questions:
1. How do you find the slope of a line given two points?
2. What is the general form of the equation of a line?
3. How do you determine if two lines are perpendicular?
4. What is the point-slope form of a line, and when is it useful?
5. How do you convert from point-slope form to slope-intercept form?

### Tip:
When writing equations for parallel lines, remember that only the intercept changes, not the slope.

Find an equation for the line that:

a) is parallel to the line defined by y = 2x + 3 and passes through the point (4, 5).

b) is parallel to the line defined by y = x + 1 and passes through the point (1, 3).

c) is perpendicular to the line defined by y = -1/2x + 2 and passes through the point (3, -5).

d) is perpendicular to the line defined by y = 1/2x - 5 and passes through the point (-2, -6).

Learn how to find equations for lines that are parallel or perpendicular to given lines and pass through specific points. This problem covers core concepts in coordinate geometry, including the point-slope form, slope-intercept form, and determining slopes for parallel and perpendicular lines. Suitable for students in Grades 9-12.

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