Question

Select the TWO lines from this list that are perpendicular.(1 point) Responses 5x−3y=−3 5 x minus 3 y is equal to negative 3 y−2=34(x+1) y minus 2 is equal to 3 fourths times open paren x plus 1 close paren 4x−3y=−6 4 x minus 3 y is equal to negative 6 y=−34x+12

Answer 1

To determine which lines are perpendicular, we need to find the slopes of the lines represented by the equations. Two lines are perpendicular if the product of their slopes is -1.

1. **First Equation: \(5x - 3y = -3\)**
- Rearranging to slope-intercept form \(y = mx + b\):
\[
3y = 5x + 3 \\
y = \frac{5}{3}x + 1
\]
- Slope \(m_1 = \frac{5}{3}\)

2. **Second Equation: \(y - 2 = \frac{3}{4}(x + 1)\)**
- Expanding and rearranging to slope-intercept form:
\[
y - 2 = \frac{3}{4}x + \frac{3}{4} \\
y = \frac{3}{4}x + \frac{11}{4}
\]
- Slope \(m_2 = \frac{3}{4}\)

3. **Third Equation: \(4x - 3y = -6\)**
- Rearranging to slope-intercept form:
\[
3y = 4x + 6 \\
y = \frac{4}{3}x + 2
\]
- Slope \(m_3 = \frac{4}{3}\)

4. **Fourth Equation: \(y = -\frac{3}{4}x + 12\)**
- This is already in slope-intercept form.
- Slope \(m_4 = -\frac{3}{4}\)

Now, let's check the slopes:
- \(m_1 = \frac{5}{3}\)
- \(m_2 = \frac{3}{4}\)
- \(m_3 = \frac{4}{3}\)
- \(m_4 = -\frac{3}{4}\)

Now we check the pairs:

- **\(m_1\) and \(m_4\)**:
\[
m_1 \cdot m_4 = \frac{5}{3} \cdot -\frac{3}{4} = -\frac{5}{4} \, \text{(not perpendicular)}
\]

- **\(m_2\) and \(m_3\)**:
\[
m_2 \cdot m_3 = \frac{3}{4} \cdot \frac{4}{3} = 1 \, \text{(not perpendicular)}
\]

- **\(m_3\) and \(m_4\)**:
\[
m_3 \cdot m_4 = \frac{4}{3} \cdot -\frac{3}{4} = -1 \, \text{(perpendicular)}
\]

Thus, the lines represented by the equations **\(4x - 3y = -6\) (third equation)** and **\(y = -\frac{3}{4}x + 12\) (fourth equation)** are perpendicular.

**Final answer:**
1. \(4x - 3y = -6\)
2. \(y = -\frac{3}{4}x + 12\)