Question

Select the TWO lines from this list that are perpendicular.(1 point) Responses y=−34x+12 y is equal to negative 3 fourths x plus 12 y−2=34(x+1) y minus 2 is equal to 3 fourths times open paren x plus 1 close paren 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

Answer 1

To determine which lines are perpendicular, we need to find the slopes of the given equations. Lines are perpendicular if the product of their slopes is -1 (the slopes are negative reciprocals of each other).

Let's rewrite the equations in slope-intercept form \(y = mx + b\), where \(m\) is the slope.

1. **\(y = -\frac{3}{4}x + 12\)**
Slope: \(m_1 = -\frac{3}{4}\)

2. **\(y - 2 = \frac{3}{4}(x + 1)\)**
Distribute:
\(y - 2 = \frac{3}{4}x + \frac{3}{4}\)
\(y = \frac{3}{4}x + \frac{3}{4} + 2\)
\(y = \frac{3}{4}x + \frac{11}{4}\)
Slope: \(m_2 = \frac{3}{4}\)

3. **\(4x - 3y = -6\)**
Rearranging:
\(-3y = -4x - 6\)
\(y = \frac{4}{3}x + 2\)
Slope: \(m_3 = \frac{4}{3}\)

4. **\(5x - 3y = -3\)**
Rearranging:
\(-3y = -5x - 3\)
\(y = \frac{5}{3}x + 1\)
Slope: \(m_4 = \frac{5}{3}\)

Now, we have the slopes:
- \(m_1 = -\frac{3}{4}\)
- \(m_2 = \frac{3}{4}\)
- \(m_3 = \frac{4}{3}\)
- \(m_4 = \frac{5}{3}\)

Next, we check for perpendicular slopes:

- The negative reciprocal of \(m_1 = -\frac{3}{4}\) is \(\frac{4}{3}\) (which is \(m_3\)).
- The negative reciprocal of \(m_2 = \frac{3}{4}\) is \(-\frac{4}{3}\) (which does not match any slope).

Thus, the lines that are perpendicular are:
1. **\(y = -\frac{3}{4}x + 12\)** (slope \(m_1 = -\frac{3}{4}\))
2. **\(4x - 3y = -6\)** (slope \(m_3 = \frac{4}{3}\))

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