For a line given by \( Ax + By = C \), the slope is:
\[ m = -\frac{A}{B} \quad (B \neq 0) \]For perpendicular lines, the product of slopes equals \(-1\):
\[ m \cdot m_{\perp} = -1 \]Hence the perpendicular slope is:
\[ m_{\perp} = -\frac{1}{m} = \frac{B}{A} \quad (A \neq 0) \]Using point-slope form:
\[ y - y_0 = m_{\perp}(x - x_0) \]If \( B = 0 \), the given line is vertical (\( x = \text{constant} \)), so the perpendicular line is horizontal through the point: \( y = y_0 \).
If \( A = 0 \), the given line is horizontal (\( y = \text{constant} \)), so the perpendicular line is vertical through the point: \( x = x_0 \).