The problem asks us to determine whether the given pair of lines are parallel, perpendicular, or neither. The equations of the lines are $-6x + 7y = 42$ and $7y + 6x = 19$.

1. Problem Description

The problem asks us to determine whether the given pair of lines are parallel, perpendicular, or neither. The equations of the lines are $-6x + 7y = 42$ and $7y + 6x = 19$.

2. Solution Steps

To determine if the lines are parallel, perpendicular, or neither, we need to find their slopes. We can rewrite each equation in slope-intercept form, $y = mx + b$, where $m$ is the slope.

For the first equation, $-6x + 7y = 42$, we can solve for $y$:

$7y = 6x + 42$

$y = \frac{6}{7}x + 6$

So the slope of the first line is $m_1 = \frac{6}{7}$.

For the second equation, $7y + 6x = 19$, we can solve for $y$:

$7y = -6x + 19$

$y = -\frac{6}{7}x + \frac{19}{7}$

So the slope of the second line is $m_2 = -\frac{6}{7}$.

Two lines are parallel if their slopes are equal, i.e., $m_1 = m_2$. In this case, $\frac{6}{7} \neq -\frac{6}{7}$, so the lines are not parallel.

Two lines are perpendicular if the product of their slopes is $-1$, i.e., $m_1 \cdot m_2 = -1$. In this case, $m_1 \cdot m_2 = (\frac{6}{7})(-\frac{6}{7}) = -\frac{36}{49}$. Since $-\frac{36}{49} \neq -1$, the lines are not perpendicular.

Therefore, the lines are neither parallel nor perpendicular.

3. Final Answer

Neither