Question

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

Answer 1

To determine which lines are perpendicular, we need to find the slopes of each line and see if the slopes are negative reciprocals of each other (i.e., the product of their slopes is -1).

Let's start by converting each line to the slope-intercept form (y = mx + b), where m is the slope.

**Line A:**
\( Y - 2 = \frac{3}{4}(x + 1) \)
Distributing:
\( Y - 2 = \frac{3}{4}x + \frac{3}{4} \)
Adding 2 to both sides:
\( Y = \frac{3}{4}x + \frac{3}{4} + 2 \)
Converting 2 to quarters: \( 2 = \frac{8}{4} \)
\( Y = \frac{3}{4}x + \frac{3}{4} + \frac{8}{4} \)
\( Y = \frac{3}{4}x + \frac{11}{4} \)
**Slope of A:** \( m_A = \frac{3}{4} \)

**Line B:**
\( 4x - 3y = -6 \)
Rearranging:
\( -3y = -4x - 6 \)
Dividing by -3:
\( y = \frac{4}{3}x + 2 \)
**Slope of B:** \( m_B = \frac{4}{3} \)

**Line C:**
\( 5x - 3y = -3 \)
Rearranging:
\( -3y = -5x - 3 \)
Dividing by -3:
\( y = \frac{5}{3}x + 1 \)
**Slope of C:** \( m_C = \frac{5}{3} \)

**Line D:**
\( Y = -\frac{3}{4}x + 12 \)
**Slope of D:** \( m_D = -\frac{3}{4} \)

Now, we can compare the slopes:

- The slopes of lines A, B, and C are positive.
- Line D has a negative slope of \(-\frac{3}{4}\).

Now let's check the product of slopes of line A and D:
- Slope of A: \( \frac{3}{4} \)
- Slope of D: \( -\frac{3}{4} \)

The product:
\( \left(\frac{3}{4}\right) \cdot \left(-\frac{3}{4}\right) = -\frac{9}{16} \)

That indicates lines A and D are not perpendicular.

Next, let's check lines B and D:
- Slope of B: \( \frac{4}{3} \)
- Slope of D: \( -\frac{3}{4} \)

The product:
\[
\left(\frac{4}{3}\right) \cdot \left(-\frac{3}{4}\right) = -1
\]

Thus, the slopes of lines B and D are negative reciprocals, making them perpendicular.

**Answer:** The two lines that are perpendicular are B and D.