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The problem asks for two equations of lines:

1. **Equation of a line parallel** to the given line \( y = \frac{2}{7}x + 3 \) that passes through the point \( (6,5) \).
2. **Equation of a line perpendicular** to the same given line, passing through the same point.

### Step 1: Find the equation of the parallel line.
The slope of the given line is \( m = \frac{2}{7} \). Since parallel lines have the same slope, the equation of the new line will also have a slope of \( \frac{2}{7} \).

We will use the point-slope form of the equation of a line:
\[
y - y_1 = m(x - x_1)
\]
where \( m = \frac{2}{7} \) and the point \( (x_1, y_1) = (6,5) \).

Substituting:
\[
y - 5 = \frac{2}{7}(x - 6)
\]
Now simplify:
\[
y - 5 = \frac{2}{7}x - \frac{12}{7}
\]
\[
y = \frac{2}{7}x - \frac{12}{7} + 5
\]
Convert 5 to a fraction with denominator 7:
\[
y = \frac{2}{7}x - \frac{12}{7} + \frac{35}{7}
\]
\[
y = \frac{2}{7}x + \frac{23}{7}
\]

So, the equation of the parallel line is:
\[
y = \frac{2}{7}x + \frac{23}{7}
\]

### Step 2: Find the equation of the perpendicular line.
For a line perpendicular to \( y = \frac{2}{7}x + 3 \), the slope will be the negative reciprocal of \( \frac{2}{7} \). The negative reciprocal of \( \frac{2}{7} \) is \( -\frac{7}{2} \).

Using the same point-slope form with \( m = -\frac{7}{2} \) and the point \( (6, 5) \), we have:
\[
y - 5 = -\frac{7}{2}(x - 6)
\]
Simplify:
\[
y - 5 = -\frac{7}{2}x + 21
\]
\[
y = -\frac{7}{2}x + 21 + 5
\]
\[
y = -\frac{7}{2}x + 26
\]

So, the equation of the perpendicular line is:
\[
y = -\frac{7}{2}x + 26
\]

### Final Answer:
- The equation of the line parallel to \( y = \frac{2}{7}x + 3 \) is \( y = \frac{2}{7}x + \frac{23}{7} \).
- The equation of the line perpendicular to \( y = \frac{2}{7}x + 3 \) is \( y = -\frac{7}{2}x + 26 \).

Would you like more details or have any questions?

### 5 Related Questions:
1. How do you find the slope of a line from an equation in standard form?
2. What is the relationship between the slopes of parallel and perpendicular lines?
3. How do you convert from point-slope form to slope-intercept form?
4. What is the general form of a linear equation?
5. How do you graph a line given its equation in slope-intercept form?

### Tip:
When working with fractions in equations, it helps to clear denominators early if possible to simplify calculations!

Write an equation for a line passing through the point (6,5) that is parallel to y = (2/7)x + 3. Then write a second equation for a line passing through the given point that is perpendicular to the given line.

Learn how to find the equation of a line passing through (6,5) parallel to y = (2/7)x + 3, and a line perpendicular to it. This problem demonstrates key concepts in algebra involving linear equations and slope relationships. Suitable for Grades 9-10 students.

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