Slope, \(m = \frac{-4}{3}\)

Perpendicular slope, \(m_2 = \frac{3}{4})

Since the line is perpendicular to \( \frac{x}{3} + \frac{y}{4} =1\) with no specified point, thus we can assume that the line will intersect the given line at \( (x_1, y_1)\)

\((y-y_1) = \frac{3}{4} (x-x_1)\)

\(y = \frac{3}{4}x - \frac{3}{4}x_1 + y_1 \)

\(4y = 3x + c\)

\(0 = 3x-4y +c\), where \(c\) is a constant value.