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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

1 answer

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.