Question

Select the TWO lines from this list that are perpendicular.(1 point) Responses 4x−3y=−64x−3y=−64 x minus 3 y is equal to negative 6 y−2=34(x+1)y−2=34(x+1)y minus 2 is equal to 3 fourths times open paren x plus 1 close paren y=−34x+12y=−34x+12y is equal to negative 3 fourths x plus 12 5x−3y=−3

Answer 1

To determine which lines are perpendicular, we first need to find the slopes of each line given in the standard form or point-slope form. Lines are perpendicular if the product of their slopes is -1.

1. **Line 1:** \( 4x - 3y = -6 \)

To find the slope, we can rewrite it in slope-intercept form (\(y = mx + b\)):
\[
-3y = -4x - 6 \\
y = \frac{4}{3}x + 2 \\
\text{Slope (m1)} = \frac{4}{3}
\]

2. **Line 2:** \( y - 2 = \frac{3}{4}(x + 1) \)

This is already in point-slope form. The slope is:
\[
\text{Slope (m2)} = \frac{3}{4}
\]

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

The slope is directly given:
\[
\text{Slope (m3)} = -\frac{3}{4}
\]

4. **Line 4:** \( 5x - 3y = -3 \)

Rewrite in slope-intercept form:
\[
-3y = -5x - 3 \\
y = \frac{5}{3}x + 1 \\
\text{Slope (m4)} = \frac{5}{3}
\]

5. **Line 5:** \( y = -\frac{3}{4}x + 12 \)

The slope is the same as for line 3:
\[
\text{Slope (m5)} = -\frac{3}{4}
\]

Now to find the pairs of lines with perpendicular slopes:

- \( \frac{4}{3} \) (line 1) and \( -\frac{3}{4} \) (line 3 or 5):
\[
\frac{4}{3} \times -\frac{3}{4} = -1 \quad \text{(perpendicular)}
\]

- \( \frac{5}{3} \) (line 4) and \( -\frac{3}{4} \) (line 3 or 5):
\[
\frac{5}{3} \times -\frac{3}{5} = -1 \quad \text{(perpendicular)}
\]

The two lines that are perpendicular are:
- **Line 1:** \( 4x - 3y = -6 \) (slope \( \frac{4}{3} \))
- **Line 3:** \( y = -\frac{3}{4}x + 12 \) (slope \( -\frac{3}{4} \))

Thus, the answer is:
- \( 4x - 3y = -6 \)
- \( y = -\frac{3}{4}x + 12 \) (or either Line 3 or Line 5)