Question

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

Answer 1

To determine which two lines are perpendicular, we need to find the slopes of each line and then check if the product of the slopes is -1 (since perpendicular lines have slopes that are negative reciprocals of each other).

1. **Line A: Y = -3/4x + 12**
- The slope is -3/4.

2. **Line B: 5x - 3y = -3**
- First, rearranging the equation into slope-intercept form (y = mx + b):
\[
-3y = -5x - 3 \\
y = \frac{5}{3}x + 1
\]
- The slope is 5/3.

3. **Line C: Y - 2 = 3/4(x + 1)**
- Expanding the equation:
\[
Y - 2 = \frac{3}{4}x + \frac{3}{4} \\
Y = \frac{3}{4}x + 2 + \frac{3}{4} = \frac{3}{4}x + \frac{11}{4}
\]
- The slope is 3/4.

4. **Line D: 4x - 3y = -6**
- Rearranging to slope-intercept form:
\[
-3y = -4x - 6 \\
y = \frac{4}{3}x + 2
\]
- The slope is 4/3.

Now we will summarize the slopes:

- Line A: slope = -3/4
- Line B: slope = 5/3
- Line C: slope = 3/4
- Line D: slope = 4/3

Now we will check for pairs of slopes that multiply to -1:

- For Line A (-3/4) and Line B (5/3):
\[
(-3/4) \times (5/3) = -15/12 \quad \text{(not equal to -1)}
\]

- For Line A (-3/4) and Line C (3/4):
\[
(-3/4) \times (3/4) = -9/16 \quad \text{(not equal to -1)}
\]

- For Line A (-3/4) and Line D (4/3):
\[
(-3/4) \times (4/3) = -12/12 = -1 \quad \text{(perpendicular)}
\]

- For Line B (5/3) and Line C (3/4):
\[
(5/3) \times (3/4) = 15/12 \quad \text{(not equal to -1)}
\]

- For Line B (5/3) and Line D (4/3):
\[
(5/3) \times (4/3) = 20/9 \quad \text{(not equal to -1)}
\]

- For Line C (3/4) and Line D (4/3):
\[
(3/4) \times (4/3) = 12/12 = 1 \quad \text{(not negative)}
\]

The only pair of lines that are perpendicular are **Line A and Line D**.

**Answer: Lines A and D are perpendicular.**