Question

Select the TWO lines from this list that are perpendicular.(1 point)
Responses

y−2=3/4(x+1)

5x−3y=−3

4x−3y=−6

y=−34x+12

Answer 1

To determine which lines are perpendicular, we need to find slopes of the given equations.

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

Rearranging it to slope-intercept form \( y = mx + b \):
\[
y - 2 = \frac{3}{4}x + \frac{3}{4} \implies y = \frac{3}{4}x + \frac{3}{4} + 2 \implies y = \frac{3}{4}x + \frac{11}{4}
\]
The slope (m) is \( \frac{3}{4} \).

2. For the second equation: \( 5x - 3y = -3 \)

Rearranging it to slope-intercept form:
\[
-3y = -5x - 3 \implies y = \frac{5}{3}x + 1
\]
The slope is \( \frac{5}{3} \).

3. For the third equation: \( 4x - 3y = -6 \)

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

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

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

Now, let's identify which lines are perpendicular. Two lines are perpendicular if the product of their slopes is -1.

- The slopes we found are:
1. \( \frac{3}{4} \)
2. \( \frac{5}{3} \)
3. \( \frac{4}{3} \)
4. \( -\frac{3}{4} \)

Now, check the products of the slopes:
- \( \frac{3}{4} \) and \( -\frac{3}{4} \):
\[
\frac{3}{4} \times -\frac{3}{4} = -\frac{9}{16} \neq -1
\]

- \( \frac{5}{3} \) and \( -\frac{3}{4} \):
\[
\frac{5}{3} \times -\frac{3}{4} = -\frac{15}{12} = -\frac{5}{4} \neq -1
\]

- \( \frac{4}{3} \) and \( -\frac{3}{4} \):
\[
\frac{4}{3} \times -\frac{3}{4} = -1
\]

Thus, the two lines that are perpendicular are:

- \( 4x - 3y = -6 \) (slope \( \frac{4}{3} \))
- \( y = -\frac{3}{4}x + 12 \) (slope \( -\frac{3}{4} \))

So the correct response is:

**4x−3y=−6** and **y=−34x+12**.