Question

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

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

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

y=−34x+12

Answer 1

To determine which lines are perpendicular, we first need to find the slope of each line. Two lines are perpendicular if the product of their slopes is -1, meaning that one slope is the negative reciprocal of the other.

1. **From the equation \(4x - 3y = -6\)**:
- Rearranging this to slope-intercept form (y = mx + b):
\[
-3y = -4x - 6 \implies y = \frac{4}{3}x + 2
\]
- Slope \(m_1 = \frac{4}{3}\).

2. **From the equation \(5x - 3y = -3\)**:
- Rearranging to slope-intercept form:
\[
-3y = -5x - 3 \implies y = \frac{5}{3}x + 1
\]
- Slope \(m_2 = \frac{5}{3}\).

3. **From the equation \(y - 2 = \frac{3}{4}(x + 1)\)**:
- This is already in point-slope form. Rearranging it:
\[
y - 2 = \frac{3}{4}x + \frac{3}{4} \implies y = \frac{3}{4}x + \frac{11}{4}
\]
- Slope \(m_3 = \frac{3}{4}\).

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

Now, let's check for perpendicular pairs:

- \(m_1 = \frac{4}{3}\) and \(m_4 = -\frac{3}{4}\):
- The product is \(\frac{4}{3} \times -\frac{3}{4} = -1\). These lines are **perpendicular**.

- \(m_2 = \frac{5}{3}\) and \(m_4 = -\frac{3}{4}\):
- The product is \(\frac{5}{3} \times -\frac{3}{4} \neq -1\). These lines are not perpendicular.

- \(m_3 = \frac{3}{4}\) and \(m_4 = -\frac{3}{4}\):
- The product is \(\frac{3}{4} \times -\frac{3}{4} \neq -1\). These lines are not perpendicular.

From this analysis, the TWO lines that are perpendicular are:

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