Theequationoflineqis

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

1 answer

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.