Question

10. Are the pairs of lines parallel, perpendicular, or intersecting?
y- 6 = 1/4 (x + 2)
y = -4x - 10

y = -2/3x + 7
-2x - 3y = 12

2x - y = 7
3x - 4y = -8

A) Intersecting
B) Perpendicular
C) Parallel

12. Select the TWO lines from this list that are perpendicular.
A) 5x − 3y = −3
B) y − 2 = 3/4 (x + 1)
C) y = −3/4x + 12
D) 4x − 3y = −6

Answer 1

To determine whether the pairs of lines are parallel, perpendicular, or intersecting, we can rewrite their equations in slope-intercept form (y = mx + b), where m is the slope.

### 10. Analyzing the Pairs of Lines:

**1. Pair 1:**
- Line 1: \( y - 6 = \frac{1}{4}(x + 2) \)
- Rearranging we have:
\[
y - 6 = \frac{1}{4}x + \frac{1}{2} \quad \Rightarrow \quad y = \frac{1}{4}x + \frac{13}{2} \quad (m_1 = \frac{1}{4})
\]

- Line 2: \( y = -4x - 10 \)
- Slope \( (m_2 = -4) \)

**Checking Slopes:**
- The product of the slopes: \( m_1 \cdot m_2 = \frac{1}{4} \cdot (-4) = -1 \).
- Since the product of the slopes is -1, these lines are **Perpendicular**.

---

**2. Pair 2:**
- Line 1: \( y = -\frac{2}{3}x + 7 \)
- Slope \( (m_1 = -\frac{2}{3}) \)

- Line 2: Convert \( -2x - 3y = 12 \) to slope-intercept form:
\[
-3y = 2x + 12 \quad \Rightarrow \quad y = -\frac{2}{3}x - 4 \quad (m_2 = -\frac{2}{3})
\]

**Checking Slopes:**
- Since both slopes are equal \( (-\frac{2}{3}) \), these lines are **Parallel**.

---

**3. Pair 3:**
- Line 1: Convert \( 2x - y = 7 \) to slope-intercept form:
\[
-y = -2x + 7 \quad \Rightarrow \quad y = 2x - 7 \quad (m_1 = 2)
\]

- Line 2: Convert \( 3x - 4y = -8 \) to slope-intercept form:
\[
-4y = -3x - 8 \quad \Rightarrow \quad y = \frac{3}{4}x + 2 \quad (m_2 = \frac{3}{4})
\]

**Checking Slopes:**
- These slopes are not equal and their product is not -1:
\[
m_1 \cdot m_2 = 2 \cdot \frac{3}{4} = \frac{3}{2}
\]
- Therefore, these lines are **Intersecting**.

### Summary for Question 10:
1. **Perpendicular** (Pair 1)
2. **Parallel** (Pair 2)
3. **Intersecting** (Pair 3)

---

### 12. Selecting TWO Lines That Are Perpendicular:

We can check the slopes of the lines provided:

**A.** \( 5x - 3y = -3 \)
- Rearranging gives \( y = \frac{5}{3}x + 1 \) (slope \( m_A = \frac{5}{3} \))

**B.** \( y - 2 = \frac{3}{4}(x + 1) \)
- Rearranging gives \( y = \frac{3}{4}x + \frac{11}{4} \) (slope \( m_B = \frac{3}{4} \))

**C.** \( y = -\frac{3}{4}x + 12 \) (slope \( m_C = -\frac{3}{4} \))

**D.** \( 4x - 3y = -6 \)
- Rearranging gives \( y = \frac{4}{3}x + 2 \) (slope \( m_D = \frac{4}{3} \))

### Checking for Perpendicular Pairs:
- \( m_B = \frac{3}{4} \) and \( m_C = -\frac{3}{4} \) (These are not perpendicular)
- \( m_A = \frac{5}{3} \) and \( m_D = \frac{4}{3} \) (Not perpendicular)
- \( m_A = \frac{5}{3} \) and \( m_C = -\frac{3}{4} \):
- \( m_A \cdot m_C = \frac{5}{3} \cdot -\frac{3}{4} = -\frac{5}{4} \) (Not perpendicular)
- \( m_B = \frac{3}{4} \) and \( m_D = \frac{4}{3} \):
- \( m_B \cdot m_D = \frac{3}{4} \cdot \frac{4}{3} = 1 \) (Not perpendicular)

#### Finding the Two Perpendicular Lines:
From the analysis,
- Line B: \( m_B = \frac{3}{4} \)
- Line C: \( m_C = -\frac{3}{4} \)

Thus, **options B and C are perpendicular**.

### Summary for Question 12:
- Lines **B** and **C** are the two lines that are **perpendicular**.