(N/A) The given line is $Ax + By + C = 0$.
Rewriting it in slope-intercept form $y = mx + c$,we get $By = -Ax - C$,which implies $y = (\frac{-A}{B})x + (\frac{-C}{B})$.
The slope of this line is $m = \frac{-A}{B}$.
Since parallel lines have equal slopes,the slope of the required line is also $m = \frac{-A}{B}$.
Using the point-slope form of a line,the equation of the line passing through $(x_{1}, y_{1})$ with slope $m$ is $y - y_{1} = m(x - x_{1})$.
Substituting $m = \frac{-A}{B}$,we get $y - y_{1} = (\frac{-A}{B})(x - x_{1})$.
Multiplying both sides by $B$,we get $B(y - y_{1}) = -A(x - x_{1})$.
Rearranging the terms,we get $A(x - x_{1}) + B(y - y_{1}) = 0$.
Thus,the equation of the line is $A(x - x_{1}) + B(y - y_{1}) = 0$.