Determine whether the pair of lines are parallel, perpendicular, or neither. \[ \left\{\begin{array}{l}x-3 y=4 \\ y=1-3 x\end{array}\right. \]

To determine the relationship between the lines defined by the equations \(x - 3y = 4\) and \(y = 1 - 3x\), we first need to put both equations into slope-intercept form \((y = mx + b)\), where \(m\) is the slope. Starting with the first equation: 1. \(x - 3y = 4\) - Rearranging gives us: \(-3y = -x + 4\) - Dividing by \(-3\): \(y = \frac{1}{3}x - \frac{4}{3}\) So, the slope of the first line is \(\frac{1}{3}\). Now, looking at the second equation: 2. \(y = 1 - 3x\) - This is already in slope-intercept form: \(y = -3x + 1\) Thus, the slope of the second line is \(-3\). Next, we analyze the slopes: - Lines are parallel if their slopes are equal. - Lines are perpendicular if the product of their slopes is \(-1\). Calculating the product of the slopes: \[\frac{1}{3} \times (-3) = -1\] Since the product of the slopes is \(-1\), the lines are perpendicular to each other.