Question

Theequationoflineqis
y
=
–78
87
x
− 11
. The
equationoflineris
y
=
–87
78
x
+
68
13
. Are
lineqandlinerparallelorperpendicular?

Answer 1

To determine whether the lines given by the equations \( q: y = -\frac{78}{87}x - 11 \) and \( r: y = -\frac{87}{78}x + \frac{68}{13} \) are parallel or perpendicular, we need to look at their slopes.

1. **Identify the slopes of the lines**:
- For line \( q \), the slope \( m_q = -\frac{78}{87} \).
- For line \( r \), the slope \( m_r = -\frac{87}{78} \).

2. **Simplify the slopes**:
- The slope of line \( q \):
\[
m_q = -\frac{78}{87} = -\frac{6}{7} \quad \text{(after dividing by 13)}
\]
- The slope of line \( r \):
\[
m_r = -\frac{87}{78} = -\frac{87}{78} = -\frac{87}{78} \quad \text{(already in simplest terms)}
\]

3. **Check if the lines are parallel**:
- Lines are parallel if their slopes are equal. Here,
\[
\text{since} \; -\frac{6}{7} \neq -\frac{87}{78}, \text{ the lines are not parallel.}
\]

4. **Check if the lines are perpendicular**:
- Lines are perpendicular if the product of their slopes equals -1:
\[
m_q \times m_r = \left(-\frac{6}{7}\right) \times \left(-\frac{87}{78}\right)
\]
- Calculate the product:
\[
= \frac{6 \times 87}{7 \times 78} = \frac{522}{546}
\]
- Simplifying \( \frac{522}{546} = \frac{87}{91} \) (dividing both numerator and denominator by 6) which is not equal to -1.

Thus, since the slopes are neither equal (not parallel) nor do they satisfy the condition for being perpendicular, we conclude:

**The lines \( q \) and \( r \) are neither parallel nor perpendicular.**