Question

Select the TWO lines from this list that are perpendicular.
A) 5x - 3y = -3
B) y - 2 = 3/4 (x + 1)
C) y = -3/4 x + 12
D) 4x - 3y = -6

Answer 1

To determine which lines are perpendicular, we need to find their slopes. Two lines are perpendicular if the product of their slopes equals -1.

Let's rewrite each of the lines in slope-intercept form (y = mx + b) or identify the slopes directly.

**A) 5x - 3y = -3**
Rearranging to slope-intercept form:
\[
-3y = -5x - 3
\]
\[
y = \frac{5}{3}x + 1
\]
The slope (m) of line A is \(\frac{5}{3}\).

---

**B) y - 2 = \frac{3}{4}(x + 1)**
Rearranging to slope-intercept form:
\[
y - 2 = \frac{3}{4}x + \frac{3}{4}
\]
\[
y = \frac{3}{4}x + \frac{11}{4}
\]
The slope (m) of line B is \(\frac{3}{4}\).

---

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

---

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

---

Now we have the slopes:
- Line A: \(\frac{5}{3}\)
- Line B: \(\frac{3}{4}\)
- Line C: \(-\frac{3}{4}\)
- Line D: \(\frac{4}{3}\)

To find perpendicular lines, we check pairs of slopes. We need m1 * m2 = -1.

1. \(\frac{5}{3}\) and \(\frac{3}{4}\):
\(\frac{5}{3} * \frac{3}{4} = \frac{15}{12} \neq -1\)

2. \(\frac{5}{3}\) and \(-\frac{3}{4}\):
\(\frac{5}{3} * -\frac{3}{4} = -\frac{15}{12} = -\frac{5}{4} \neq -1\)

3. \(\frac{5}{3}\) and \(\frac{4}{3}\):
\(\frac{5}{3} * \frac{4}{3} = \frac{20}{9} \neq -1\)

4. \(\frac{3}{4}\) and \(-\frac{3}{4}\):
\(\frac{3}{4} * -\frac{3}{4} = -\frac{9}{16} \neq -1\)

5. \(\frac{3}{4}\) and \(\frac{4}{3}\):
\(\frac{3}{4} * \frac{4}{3} = 1 \neq -1\)

6. \(-\frac{3}{4}\) and \(\frac{4}{3}\):
\(-\frac{3}{4} * \frac{4}{3} = -1\)

Thus, lines C (\(y = -\frac{3}{4}x + 12\)) and D (\(4x - 3y = -6\)) are perpendicular.

**Final Answer:** C) \(y = -\frac{3}{4}x + 12\) and D) \(4x - 3y = -6\) are perpendicular.