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To solve this problem, we need to find the equation of lines that are perpendicular to the given lines and pass through the specific points provided.

### Key steps:
1. **Find the slope of each given line.** The slope \( m_1 \) of a line in slope-intercept form \( y = mx + b \) is the coefficient of \( x \). 
2. **Calculate the slope of the perpendicular line.** For a line to be perpendicular, its slope \( m_2 \) must satisfy \( m_1 \cdot m_2 = -1 \), or \( m_2 = -\frac{1}{m_1} \).
3. **Use the point-slope form** \( y - y_1 = m(x - x_1) \) to find the equation of the line passing through the given point \( (x_1, y_1) \).
4. **Convert to slope-intercept form** \( y = mx + b \).

Let’s go through each equation:

### 1. \( y = 3x + 6 \) ; \( (6,7) \)
- Slope of given line: \( m_1 = 3 \)
- Slope of perpendicular line: \( m_2 = -\frac{1}{3} \)
- Use point \( (6, 7) \):
  \[
  y - 7 = -\frac{1}{3}(x - 6)
  \]
  \[
  y - 7 = -\frac{1}{3}x + 2
  \]
  \[
  y = -\frac{1}{3}x + 9
  \]

### 2. \( y = x - 4 \) ; \( (-2, 3) \)
- Slope of given line: \( m_1 = 1 \)
- Slope of perpendicular line: \( m_2 = -1 \)
- Use point \( (-2, 3) \):
  \[
  y - 3 = -1(x + 2)
  \]
  \[
  y - 3 = -x - 2
  \]
  \[
  y = -x + 1
  \]

### 3. \( y = \frac{1}{2}x + 5 \) ; \( (4, -5) \)
- Slope of given line: \( m_1 = \frac{1}{2} \)
- Slope of perpendicular line: \( m_2 = -2 \)
- Use point \( (4, -5) \):
  \[
  y + 5 = -2(x - 4)
  \]
  \[
  y + 5 = -2x + 8
  \]
  \[
  y = -2x + 3
  \]

### 4. \( y = -2x + 4 \) ; \( (-4, 2) \)
- Slope of given line: \( m_1 = -2 \)
- Slope of perpendicular line: \( m_2 = \frac{1}{2} \)
- Use point \( (-4, 2) \):
  \[
  y - 2 = \frac{1}{2}(x + 4)
  \]
  \[
  y - 2 = \frac{1}{2}x + 2
  \]
  \[
  y = \frac{1}{2}x + 4
  \]

### 5. \( 3x - 4y = 16 \) ; \( (3,5) \)
- Rewrite the equation in slope-intercept form:
  \[
  -4y = -3x + 16 \quad \Rightarrow \quad y = \frac{3}{4}x - 4
  \]
  - Slope of given line: \( m_1 = \frac{3}{4} \)
  - Slope of perpendicular line: \( m_2 = -\frac{4}{3} \)
  - Use point \( (3, 5) \):
  \[
  y - 5 = -\frac{4}{3}(x - 3)
  \]
  \[
  y - 5 = -\frac{4}{3}x + 4
  \]
  \[
  y = -\frac{4}{3}x + 9
  \]

### Summary of Perpendicular Lines:
1. \( y = -\frac{1}{3}x + 9 \) for \( y = 3x + 6 \)
2. \( y = -x + 1 \) for \( y = x - 4 \)
3. \( y = -2x + 3 \) for \( y = \frac{1}{2}x + 5 \)
4. \( y = \frac{1}{2}x + 4 \) for \( y = -2x + 4 \)
5. \( y = -\frac{4}{3}x + 9 \) for \( 3x - 4y = 16 \)

Do you want more details or have any questions?

### Related questions:
1. How do you find the slope of a line from its equation?
2. What does it mean for two lines to be perpendicular?
3. How do you find the equation of a line given its slope and a point on the line?
4. What is the slope-intercept form of a linear equation?
5. How do you convert a linear equation from standard form to slope-intercept form?

**Tip:** The slopes of perpendicular lines are negative reciprocals of each other.

Match the equation of a line in slope-intercept form that is perpendicular to the graph of each equation and passes through the given point?

Learn how to find the equation of a line that is perpendicular to another and passes through a given point. This problem covers key algebra concepts, such as slope-intercept form, perpendicular line slopes, and point-slope form, with examples for Grade 8-10 students.

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