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

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: $y = 4x + 7$.

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: $y = -\frac{1}{2}x - 3$.

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.

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?

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: $y = -4x + 13$.

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.