Question

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

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

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−2=3/4(x+1)

Answer 1

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

1. The first line is \( y = -\frac{3}{4}x + 12 \). The slope (m) is \( -\frac{3}{4} \).

2. The second line, \( 5x - 3y = -3 \), can be rewritten in slope-intercept form (y = mx + b):
\[
-3y = -5x - 3 \implies y = \frac{5}{3}x + 1
\]
The slope here is \( \frac{5}{3} \).

3. The third line, \( 4x - 3y = -6 \), can also be rewritten in slope-intercept form:
\[
-3y = -4x - 6 \implies y = \frac{4}{3}x + 2
\]
The slope is \( \frac{4}{3} \).

4. The fourth line, \( y - 2 = \frac{3}{4}(x + 1) \), can be rearranged:
\[
y - 2 = \frac{3}{4}x + \frac{3}{4} \implies y = \frac{3}{4}x + \frac{11}{4}
\]
The slope is \( \frac{3}{4} \).

Now we summarize the slopes:
- First line: \( m_1 = -\frac{3}{4} \)
- Second line: \( m_2 = \frac{5}{3} \)
- Third line: \( m_3 = \frac{4}{3} \)
- Fourth line: \( m_4 = \frac{3}{4} \)

Now we check for perpendicularity:
- The slope \( -\frac{3}{4} \) (First line) and \( \frac{4}{3} \) (Third line) satisfy the condition for perpendicular lines because:
\[
-\frac{3}{4} \times \frac{4}{3} = -1
\]

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