https://file.aiquickdraw.com/mathbot/file/1725511534532p7g4tmrm.jpg

Let's work through the problems visible in the image.

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### Problem 18:
Write the equation of a line that:
- Passes through (0, 7) and
- Is parallel to \( y = 4x - 19 \).

**Solution**:
Since the line is parallel to \( y = 4x - 19 \), it has the same slope, \( m = 4 \).
We can use the point-slope form \( y - y_1 = m(x - x_1) \), with the point \( (0, 7) \):
\[
y - 7 = 4(x - 0)
\]
Simplifying, we get:
\[
y - 7 = 4x \implies y = 4x + 7
\]

The equation is: **\( y = 4x + 7 \)**.

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### Problem 19:
Write the equation of a line that:
- Passes through \((-8, 1)\) and
- Is perpendicular to \( y = 2x - 17 \).

**Solution**:
The slope of the line perpendicular to \( y = 2x - 17 \) is the negative reciprocal of 2, which is \( -\frac{1}{2} \).

Using point-slope form \( y - y_1 = m(x - x_1) \), with the point \( (-8, 1) \) and slope \( m = -\frac{1}{2} \):
\[
y - 1 = -\frac{1}{2}(x + 8)
\]
Simplifying:
\[
y - 1 = -\frac{1}{2}x - 4 \implies y = -\frac{1}{2}x - 3
\]

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

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### Problem 20:
Find line \( t \) so that \( t \parallel \ell \). Identify the postulate or theorem.

This problem shows a geometric figure but the details are unclear. Based on the general setup, the **parallel postulate** would likely apply here, stating that for a line and a point not on the line, there exists exactly one line through the point parallel to the given line.

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### Problem 21:
Find the distance from line \( t \) to points \( P(1, 2) \) and \( Q(-3, -5) \).

This involves the distance formula and possibly more detail from the image's diagram. Would you like to clarify or upload a more zoomed-in portion for accuracy?

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### Problem 22:
Write an equation for the perpendicular bisector of \( AB \) where \( A = (-2, 4) \) and \( B = (6, 6) \).

The perpendicular bisector will pass through the midpoint of \( A \) and \( B \) and have a slope that is the negative reciprocal of the slope of \( AB \).

**Step 1**: Find the midpoint of \( AB \):
\[
\text{Midpoint} = \left( \frac{-2 + 6}{2}, \frac{4 + 6}{2} \right) = (2, 5)
\]

**Step 2**: Find the slope of \( AB \):
\[
m_{AB} = \frac{6 - 4}{6 - (-2)} = \frac{2}{8} = \frac{1}{4}
\]

The slope of the perpendicular bisector is \( -4 \).

**Step 3**: Use point-slope form \( y - y_1 = m(x - x_1) \) with the point \( (2, 5) \) and slope \( -4 \):
\[
y - 5 = -4(x - 2)
\]
Simplifying:
\[
y - 5 = -4x + 8 \implies y = -4x + 13
\]

The equation is: **\( y = -4x + 13 \)**.

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Would you like more details or have any specific questions about these problems?

### Related Questions:
1. How do you derive the slope from a given equation of a line?
2. What are the conditions for two lines to be parallel?
3. How is the distance between a point and a line calculated geometrically?
4. What is the negative reciprocal, and why is it used for perpendicular lines?
5. How do you find the midpoint of a line segment?

**Tip**: When working with perpendicular bisectors, always double-check the negative reciprocal for the slope and ensure you're using the midpoint correctly.

Write equations in slope-intercept form for various lines based on given conditions (parallel, perpendicular, etc.), and find distances between points and lines.

Learn how to write equations in slope-intercept form for lines parallel or perpendicular to given equations, and calculate distances from points to lines. This includes understanding slope, midpoint, and perpendicular bisectors. Suitable for high school students in Grades 9-10.

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