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=−34x+12 y is equal to negative 3 fourths x plus 12 4x−3y=−6 4 x minus 3 y is equal to negative 6 y−2=34(x+1)

Answer 1

To determine which two lines are perpendicular, we need to compare their slopes. Lines are perpendicular if the product of their slopes is -1.

1. **For the first equation:** \(5x - 3y = -3\)
- Rearranging into slope-intercept form (y = mx + b):
\[
-3y = -5x - 3 \implies y = \frac{5}{3}x + 1
\]
- The slope (m) is \(\frac{5}{3}\).

2. **For the second equation:** \(y = -\frac{3}{4}x + 12\)
- The slope is \(-\frac{3}{4}\).

3. **For the third equation:** \(4x - 3y = -6\)
- Rearranging into slope-intercept form:
\[
-3y = -4x - 6 \implies y = \frac{4}{3}x + 2
\]
- The slope is \(\frac{4}{3}\).

4. **For the fourth equation:** \(y - 2 = \frac{3}{4}(x + 1)\)
- Rearranging into slope-intercept form:
\[
y - 2 = \frac{3}{4}x + \frac{3}{4} \implies y = \frac{3}{4}x + \frac{11}{4}
\]
- The slope is \(\frac{3}{4}\).

Now we summarize the slopes:
- First line (5x - 3y = -3): \(m_1 = \frac{5}{3}\)
- Second line (y = -34x + 12): \(m_2 = -\frac{3}{4}\)
- Third line (4x - 3y = -6): \(m_3 = \frac{4}{3}\)
- Fourth line (y - 2 = 34(x + 1)): \(m_4 = \frac{3}{4}\)

Next, we check for perpendicularity:
- The line with slope \(\frac{5}{3}\) and the line with slope \(-\frac{3}{4}\):
\[
\left(\frac{5}{3}\right) \cdot \left(-\frac{3}{4}\right) = -\frac{15}{12} = -\frac{5}{4}
\]
(Not perpendicular)

- The line with slope \(-\frac{3}{4}\) and the line with slope \(\frac{4}{3}\):
\[
\left(-\frac{3}{4}\right) \cdot \left(\frac{4}{3}\right) = -1
\]
(These lines are perpendicular)

Thus, the two lines that are perpendicular are:
- **(2)** \( y = -\frac{3}{4}x + 12\)
- **(3)** \( 4x - 3y = -6\)

So, the answer is the second line and the third line.