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

4x−3y=−6
4 x minus 3 y is equal to negative 6

y = −3/4 x + 12
y is equal to negative 3 fourths x plus 12

y − 2 = 3/4 (x+1)

Answer 1

To determine which two lines are perpendicular, we need to find the slopes of the given equations. Two lines are perpendicular if the product of their slopes is -1 (i.e., their slopes are negative reciprocals).

1. **For the first equation**:
\( 5x - 3y = -3 \)

Rearranging to 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**:
\( 4x - 3y = -6 \)

Rearranging:
\[
-3y = -4x - 6 \implies y = \frac{4}{3}x + 2
\]
The slope is \(\frac{4}{3}\).

3. **For the third equation**:
\( y = -\frac{3}{4}x + 12 \)

The slope is \(-\frac{3}{4}\).

4. **For the fourth equation**:
\( y - 2 = \frac{3}{4}(x + 1) \)

Expanding this:
\[
y - 2 = \frac{3}{4}x + \frac{3}{4} \implies y = \frac{3}{4}x + \frac{11}{4}
\]
The slope is \(\frac{3}{4}\).

Now, we can analyze the slopes:

1. First equation: \(m = \frac{5}{3}\)
2. Second equation: \(m = \frac{4}{3}\)
3. Third equation: \(m = -\frac{3}{4}\)
4. Fourth equation: \(m = \frac{3}{4}\)

The only pair of slopes that are negative reciprocals (i.e., their product is -1) are:

- The third equation with slope \(m = -\frac{3}{4}\)
- The first equation with slope \(m = \frac{4}{3}\) (since \(\frac{4}{3} \times -\frac{3}{4} = -1\))

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

So the response is:
- **5x−3y=−3**
- **y = −3/4 x + 12**