Coordinate Geometry worksheet with problems on slope-intercept and point-slope forms.

Problem Analysis:

The task involves solving problems in coordinate geometry using the slope-intercept form and point-slope form of linear equations. The goal is to:
1. Determine the slope of given lines.
2. Write equations for lines that are parallel or perpendicular to given lines.
3. Determine whether pairs of lines are parallel, perpendicular, or neither.
4. Write equations in diagonal intercept form.

Let's solve each part step by step.

---

Part 1: What is the slope of the following equations?

#### Equation 1: $ y = -6x + 2 $
- This equation is already in slope-intercept form: $ y = mx + b $, where $ m $ is the slope.
- Here, $ m = -6 $.
- Slope: $ \boxed{-6} $

#### Equation 2: $ 2x + y = 2 $
- Rearrange into slope-intercept form ($ y = mx + b $):
$$
y = -2x + 2
$$
- Here, $ m = -2 $.
- Slope: $ \boxed{-2} $

#### Equation 3: $ 3x + 4y = 9 $
- Rearrange into slope-intercept form:
$$
4y = -3x + 9 \implies y = -\frac{3}{4}x + \frac{9}{4}
$$
- Here, $ m = -\frac{3}{4} $.
- Slope: $ \boxed{-\frac{3}{4}} $

#### Equation 4: $ 4x - 2y = 3 $
- Rearrange into slope-intercept form:
$$
-2y = -4x + 3 \implies y = 2x - \frac{3}{2}
$$
- Here, $ m = 2 $.
- Slope: $ \boxed{2} $

#### Equation 5: $ 5x - 2y = 10 $
- Rearrange into slope-intercept form:
$$
-2y = -5x + 10 \implies y = \frac{5}{2}x - 5
$$
- Here, $ m = \frac{5}{2} $.
- Slope: $ \boxed{\frac{5}{2}} $

#### Equation 6: $ -3x + y = 8 $
- Rearrange into slope-intercept form:
$$
y = 3x + 8
$$
- Here, $ m = 3 $.
- Slope: $ \boxed{3} $

---

Part 2: Write the equation for each line that is parallel to the given line.

#### Line 1: $ y = 5x - 1 $; Point: $ (2, 0) $
- Slope of the given line: $ m = 5 $.
- A line parallel to this line will have the same slope, $ m = 5 $.
- Use the point-slope form: $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) = (2, 0) $:
$$
y - 0 = 5(x - 2) \implies y = 5x - 10
$$
- Equation: $ \boxed{y = 5x - 10} $

#### Line 2: $ 3x + y = 0 $; Point: $ (1, 5) $
- Rearrange the given line into slope-intercept form:
$$
y = -3x
$$
- Slope of the given line: $ m = -3 $.
- A line parallel to this line will have the same slope, $ m = -3 $.
- Use the point-slope form: $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) = (1, 5) $:
$$
y - 5 = -3(x - 1) \implies y - 5 = -3x + 3 \implies y = -3x + 8
$$
- Equation: $ \boxed{y = -3x + 8} $

#### Line 3: $ y = -1/2x + 3 $; Point: $ (3, 7) $
- Slope of the given line: $ m = -\frac{1}{2} $.
- A line parallel to this line will have the same slope, $ m = -\frac{1}{2} $.
- Use the point-slope form: $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) = (3, 7) $:
$$
y - 7 = -\frac{1}{2}(x - 3) \implies y - 7 = -\frac{1}{2}x + \frac{3}{2} \implies y = -\frac{1}{2}x + \frac{17}{2}
$$
- Equation: $ \boxed{y = -\frac{1}{2}x + \frac{17}{2}} $

#### Line 4: $ 4y = 2x + 8 $; Point: $ (0, 0) $
- Rearrange the given line into slope-intercept form:
$$
y = \frac{1}{2}x + 2
$$
- Slope of the given line: $ m = \frac{1}{2} $.
- A line parallel to this line will have the same slope, $ m = \frac{1}{2} $.
- Use the point-slope form: $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) = (0, 0) $:
$$
y - 0 = \frac{1}{2}(x - 0) \implies y = \frac{1}{2}x
$$
- Equation: $ \boxed{y = \frac{1}{2}x} $

---

Part 3: Write the equation for each line that is perpendicular to the given line.

#### Line 1: $ y = 1/3x + 4 $; Point: $ (-2, 3) $
- Slope of the given line: $ m = \frac{1}{3} $.
- Slope of a line perpendicular to this line: $ m_{\perp} = -\frac{1}{m} = -3 $.
- Use the point-slope form: $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) = (-2, 3) $:
$$
y - 3 = -3(x + 2) \implies y - 3 = -3x - 6 \implies y = -3x - 3
$$
- Equation: $ \boxed{y = -3x - 3} $

#### Line 2: $ 2x + y = 6 $; Point: $ (0, 0) $
- Rearrange the given line into slope-intercept form:
$$
y = -2x + 6
$$
- Slope of the given line: $ m = -2 $.
- Slope of a line perpendicular to this line: $ m_{\perp} = -\frac{1}{m} = \frac{1}{2} $.
- Use the point-slope form: $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) = (0, 0) $:
$$
y - 0 = \frac{1}{2}(x - 0) \implies y = \frac{1}{2}x
$$
- Equation: $ \boxed{y = \frac{1}{2}x} $

#### Line 3: $ 2x + 3y = 2 $; Point: $ (3, 0) $
- Rearrange the given line into slope-intercept form:
$$
3y = -2x + 2 \implies y = -\frac{2}{3}x + \frac{2}{3}
$$
- Slope of the given line: $ m = -\frac{2}{3} $.
- Slope of a line perpendicular to this line: $ m_{\perp} = -\frac{1}{m} = \frac{3}{2} $.
- Use the point-slope form: $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) = (3, 0) $:
$$
y - 0 = \frac{3}{2}(x - 3) \implies y = \frac{3}{2}x - \frac{9}{2}
$$
- Equation: $ \boxed{y = \frac{3}{2}x - \frac{9}{2}} $

#### Line 4: $ 4x + 2y = 5 $; Point: $ (2, -3) $
- Rearrange the given line into slope-intercept form:
$$
2y = -4x + 5 \implies y = -2x + \frac{5}{2}
$$
- Slope of the given line: $ m = -2 $.
- Slope of a line perpendicular to this line: $ m_{\perp} = -\frac{1}{m} = \frac{1}{2} $.
- Use the point-slope form: $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) = (2, -3) $:
$$
y + 3 = \frac{1}{2}(x - 2) \implies y + 3 = \frac{1}{2}x - 1 \implies y = \frac{1}{2}x - 4
$$
- Equation: $ \boxed{y = \frac{1}{2}x - 4} $

---

Part 4: Determine whether the lines are parallel, perpendicular, or neither.

#### Pair 1: $ 2y = 2x + 8 $ and $ y = 1/2x + 1 $
- First line: $ 2y = 2x + 8 \implies y = x + 4 $. Slope: $ m_1 = 1 $.
- Second line: $ y = \frac{1}{2}x + 1 $. Slope: $ m_2 = \frac{1}{2} $.
- Since $ m_1 \neq m_2 $, the lines are not parallel.
- Since $ m_1 \cdot m_2 \neq -1 $ ($ 1 \cdot \frac{1}{2} \neq -1 $), the lines are not perpendicular.
- Conclusion: Neither.

#### Pair 2: $ 3y = 2x + 3 $ and $ y = 5x + 1 $
- First line: $ 3y = 2x + 3 \implies y = \frac{2}{3}x + 1 $. Slope: $ m_1 = \frac{2}{3} $.
- Second line: $ y = 5x + 1 $. Slope: $ m_2 = 5 $.
- Since $ m_1 \neq m_2 $, the lines are not parallel.
- Since $ m_1 \cdot m_2 \neq -1 $ ($ \frac{2}{3} \cdot 5 \neq -1 $), the lines are not perpendicular.
- Conclusion: Neither.

#### Pair 3: $ Y = 2/3x + 1 $ and $ y = 3/2x + 3 $
- First line: $ Y = \frac{2}{3}x + 1 $. Slope: $ m_1 = \frac{2}{3} $.
- Second line: $ y = \frac{3}{2}x + 3 $. Slope: $ m_2 = \frac{3}{2} $.
- Since $ m_1 \neq m_2 $, the lines are not parallel.
- Since $ m_1 \cdot m_2 = \frac{2}{3} \cdot \frac{3}{2} = 1 \neq -1 $, the lines are not perpendicular.
- Conclusion: Neither.

#### Pair 4: $ Y = 1/5x + 2 $ and $ y = 5x + 1 $
- First line: $ Y = \frac{1}{5}x + 2 $. Slope: $ m_1 = \frac{1}{5} $.
- Second line: $ y = 5x + 1 $. Slope: $ m_2 = 5 $.
- Since $ m_1 \neq m_2 $, the lines are not parallel.
- Since $ m_1 \cdot m_2 = \frac{1}{5} \cdot 5 = 1 \neq -1 $, the lines are not perpendicular.
- Conclusion: Neither.

---

Part 5: Write the equations in diagonal intercept form.

Diagonal intercept form is given by:
$$
\frac{x}{a} + \frac{y}{b} = 1
$$
where $ a $ is the x-intercept and $ b $ is the y-intercept.

#### Equation 1: $ x + 2y = 3 $
- Find the x-intercept by setting $ y = 0 $:
$$
x + 2(0) = 3 \implies x = 3 \quad (\text{x-intercept} = 3)
$$
- Find the y-intercept by setting $ x = 0 $:
$$
0 + 2y = 3 \implies y = \frac{3}{2} \quad (\text{y-intercept} = \frac{3}{2})
$$
- Diagonal intercept form:
$$
\frac{x}{3} + \frac{y}{\frac{3}{2}} = 1 \implies \frac{x}{3} + \frac{2y}{3} = 1
$$
- Equation: $ \boxed{\frac{x}{3} + \frac{2y}{3} = 1} $

#### Equation 2: $ 2x + y = 12 $
- Find the x-intercept by setting $ y = 0 $:
$$
2x + 0 = 12 \implies x = 6 \quad (\text{x-intercept} = 6)
$$
- Find the y-intercept by setting $ x = 0 $:
$$
2(0) + y = 12 \implies y = 12 \quad (\text{y-intercept} = 12)
$$
- Diagonal intercept form:
$$
\frac{x}{6} + \frac{y}{12} = 1
$$
- Equation: $ \boxed{\frac{x}{6} + \frac{y}{12} = 1} $

#### Equation 3: $ 3y + 5x = 2 $
- Find the x-intercept by setting $ y = 0 $:
$$
3(0) + 5x = 2 \implies x = \frac{2}{5} \quad (\text{x-intercept} = \frac{2}{5})
$$
- Find the y-intercept by setting $ x = 0 $:
$$
3y + 5(0) = 2 \implies y = \frac{2}{3} \quad (\text{y-intercept} = \frac{2}{3})
$$
- Diagonal intercept form:
$$
\frac{x}{\frac{2}{5}} + \frac{y}{\frac{2}{3}} = 1 \implies \frac{5x}{2} + \frac{3y}{2} = 1
$$
- Equation: $ \boxed{\frac{5x}{2} + \frac{3y}{2} = 1} $

#### Equation 4: $ 6x + y = 1 $
- Find the x-intercept by setting $ y = 0 $:
$$
6x + 0 = 1 \implies x = \frac{1}{6} \quad (\text{x-intercept} = \frac{1}{6})
$$
- Find the y-intercept by setting $ x = 0 $:
$$
6(0) + y = 1 \implies y = 1 \quad (\text{y-intercept} = 1)
$$
- Diagonal intercept form:
$$
\frac{x}{\frac{1}{6}} + \frac{y}{1} = 1 \implies 6x + y = 1
$$
- Equation: $ \boxed{6x + y = 1} $

#### Equation 5: $ y = -7x $
- Find the x-intercept by setting $ y = 0 $:
$$
0 = -7x \implies x = 0 \quad (\text{x-intercept does not exist})
$$
- Find the y-intercept by setting $ x = 0 $:
$$
y = -7(0) \implies y = 0 \quad (\text{y-intercept does not exist})
$$
- Since both intercepts are zero, this line cannot be expressed in diagonal intercept form.
- Equation: $ \boxed{\text{Not applicable}} $

#### Equation 6: $ 6x + 2y = 2 $
- Find the x-intercept by setting $ y = 0 $:
$$
6x + 2(0) = 2 \implies x = \frac{1}{3} \quad (\text{x-intercept} = \frac{1}{3})
$$
- Find the y-intercept by setting $ x = 0 $:
$$
6(0) + 2y = 2 \implies y = 1 \quad (\text{y-intercept} = 1)
$$
- Diagonal intercept form:
$$
\frac{x}{\frac{1}{3}} + \frac{y}{1} = 1 \implies 3x + y = 1
$$
- Equation: $ \boxed{3x + y = 1} $

---

Final Answers:

1. Slopes:
- $ \boxed{-6} $
- $ \boxed{-2} $
- $ \boxed{-\frac{3}{4}} $
- $ \boxed{2} $
- $ \boxed{\frac{5}{2}} $
- $ \boxed{3} $

2. Parallel lines:
- $ \boxed{y = 5x - 10} $
- $ \boxed{y = -3x + 8} $
- $ \boxed{y = -\frac{1}{2}x + \frac{17}{2}} $
- $ \boxed{y = \frac{1}{2}x} $

3. Perpendicular lines:
- $ \boxed{y = -3x - 3} $
- $ \boxed{y = \frac{1}{2}x} $
- $ \boxed{y = \frac{3}{2}x - \frac{9}{2}} $
- $ \boxed{y = \frac{1}{2}x - 4} $

4. Parallel/perpendicular determination:
- Neither
- Neither
- Neither
- Neither

5. Diagonal intercept forms:
- $ \boxed{\frac{x}{3} + \frac{2y}{3} = 1} $
- $ \boxed{\frac{x}{6} + \frac{y}{12} = 1} $
- $ \boxed{\frac{5x}{2} + \frac{3y}{2} = 1} $
- $ \boxed{6x + y = 1} $
- Not applicable
- $ \boxed{3x + y = 1} $