## Thomas' Calculus 13th Edition

Solving both equations for y (rewriting in slope-intercept form) we find the slopes: $\left[\begin{array}{lllll} Ax+By=C_{1} & & ... & Bx-Ay=C_{2} & \\ By=-Ax+C_{1} & & & -Ay=-Bx=C_{2} & \\ y=-\frac{A}{B}x+\frac{C_{1}}{B} & & & y=\frac{B}{A}x-\frac{C_{2}}{A} & \\ & & & & \\ m_{1}=-\frac{A}{B} & & & m_{2}=\frac{B}{A} & \end{array}\right]$ Both slopes are defined, as $A\neq 0$ and $B\neq 0$, and we note that $m_{1}\cdot m_{2}=-1$, meaning that the two lines are perpendicular.