Find the line through (2, -3) perpendicular to 3x - 4y + 7 = 0…

The general equation of a line is $Ax + By + C = 0$. Its slope is $-\dfrac{A}{B}$, its $x$-intercept $-\dfrac{C}{A}$ and its $y$-intercept $-\dfrac{C}{B}$. Parallel to it: $Ax + By + k = 0$. Perpendicular to it: $Bx - Ay + k = 0$. The normal form is $x\cos\alpha + y\sin\alpha = p$, found by writing $Ax + By = -C$, making the right side positive and dividing by $\sqrt{A^2 + B^2}$.