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### Problem 65:
You are given two points for line \( L_1 \): \( (3, 6) \) and \( (-6, 0) \), and for line \( L_2 \): \( (0, -1) \) and \( \left( 5, \frac{7}{3} \right) \).

#### **Line \( L_1 \):**
To find the equation of the line in slope-intercept form, we first need to calculate the slope \( m \).

The slope formula is:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

For \( L_1 \):
\[
m = \frac{0 - 6}{-6 - 3} = \frac{-6}{-9} = \frac{2}{3}
\]

Using the slope \( m = \frac{2}{3} \), we now use the point-slope form to write the equation of the line. Choose the point \( (3, 6) \):
\[
y - 6 = \frac{2}{3}(x - 3)
\]
Simplifying to slope-intercept form:
\[
y - 6 = \frac{2}{3}x - 2 \quad \Rightarrow \quad y = \frac{2}{3}x + 4
\]

Thus, the equation of \( L_1 \) is:
\[
y = \frac{2}{3}x + 4
\]

#### **Line \( L_2 \):**
For line \( L_2 \), we calculate the slope using the points \( (0, -1) \) and \( \left( 5, \frac{7}{3} \right) \).

\[
m = \frac{\frac{7}{3} - (-1)}{5 - 0} = \frac{\frac{7}{3} + 1}{5} = \frac{\frac{7}{3} + \frac{3}{3}}{5} = \frac{\frac{10}{3}}{5} = \frac{10}{15} = \frac{2}{3}
\]

Since the slope is also \( \frac{2}{3} \), and we know the point \( (0, -1) \), we can directly use the slope-intercept form \( y = mx + b \). Here, \( b = -1 \) (the y-intercept is \( -1 \)).

Thus, the equation of \( L_2 \) is:
\[
y = \frac{2}{3}x - 1
\]

---

### Problem 66:
For line \( L_1 \), you are given the points \( (4, 8) \) and \( (-4, 2) \), and for line \( L_2 \), the points are \( (3, -5) \) and \( \left( -1, \frac{-1}{3} \right) \).

#### **Line \( L_1 \):**
First, calculate the slope for \( L_1 \):
\[
m = \frac{2 - 8}{-4 - 4} = \frac{-6}{-8} = \frac{3}{4}
\]

Using the point-slope form and the point \( (4, 8) \):
\[
y - 8 = \frac{3}{4}(x - 4)
\]
Simplifying:
\[
y - 8 = \frac{3}{4}x - 3 \quad \Rightarrow \quad y = \frac{3}{4}x + 5
\]

Thus, the equation of \( L_1 \) is:
\[
y = \frac{3}{4}x + 5
\]

#### **Line \( L_2 \):**
Now, calculate the slope for \( L_2 \) using \( (3, -5) \) and \( \left( -1, \frac{-1}{3} \right) \):
\[
m = \frac{\frac{-1}{3} - (-5)}{-1 - 3} = \frac{\frac{-1}{3} + 5}{-4} = \frac{\frac{-1}{3} + \frac{15}{3}}{-4} = \frac{\frac{14}{3}}{-4} = \frac{-14}{12} = \frac{-7}{6}
\]

Using the point \( (3, -5) \) and the slope \( m = \frac{-7}{6} \), we apply the point-slope form:
\[
y + 5 = \frac{-7}{6}(x - 3)
\]
Simplifying:
\[
y + 5 = \frac{-7}{6}x + \frac{7}{2} \quad \Rightarrow \quad y = \frac{-7}{6}x - \frac{3}{2}
\]

Thus, the equation of \( L_2 \) is:
\[
y = \frac{-7}{6}x - \frac{3}{2}
\]

---

Would you like to go over the parallel and perpendicular lines problem next? Let me know if you need more details on any of these!

Here are some related questions for practice:
1. How do you find the slope of a line given two points?
2. What is the point-slope form of a line, and when do you use it?
3. How do you convert point-slope form to slope-intercept form?
4. How do you find the equation of a line given its slope and a point on the line?
5. How can you determine if two lines are parallel or perpendicular?

**Tip:** When solving for the slope between two points, always be careful with the order of the coordinates to avoid sign errors.

solve and show me the work for my homework

Learn how to solve linear equations and find the equations of parallel and perpendicular lines using the slope formula and point-slope form. This problem involves calculating slopes and writing equations in slope-intercept form. Suitable for high school students in Grades 9-12.

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