Given the four equations:
$\textbf{(1)}\ 3y-2x=12 \qquad\textbf{(2)}\ -2x-3y=10 \qquad\textbf{(3)}\ 3y+2x=12 \qquad\textbf{(4)}\ 2y+3x=10$
The pair representing the perpendicular lines is:

$\textbf{(A)}\ \text{(1) and (4)}\qquad \textbf{(B)}\ \text{(1) and (3)}\qquad \textbf{(C)}\ \text{(1) and (2)}\qquad \textbf{(D)}\ \text{(2) and (4)}\qquad \textbf{(E)}\ \text{(2) and (3)}$
Write each equation in slope-intercept from since the slopes are easier to compare.
Equation $(1)$ in slope-intercept form is $y = \frac{2}{3}x+4$.
Equation $(2)$ in slope-intercept form is $y = -\frac{2}{3}x - \frac{10}{3}$.
Equation $(3)$ in slope-intercept form is $y = -\frac{2}{3}x + 4$.
Equation $(4)$ in slope-intercept form is $y = -\frac{3}{2}x + 5$.
Remember that if the two lines are perpendicular, then the product of two slopes equals $-1$.  Equations $(1)$ and $(4)$ satisfy the condition, so the answer is $\boxed{\textbf{(A)}}$.