Find k, if the line 2x - 3y = 11 is perpendicular to the line 3x + ky = -4?

$$\eqalign{ & 2x - 3y = 11 \cr & 3y = 2x - 11 \cr & y = \frac{2}{3}x - \frac{{11}}{3} \cr & \left( {y = mx + c,{\text{ where }}m{\text{ is slope}}} \right) \cr & {\text{Slope}} = {m_1} = \frac{2}{3} \cr & 3x + ky = - 4 \cr & ky = - 3x - 4 \cr & y = - \frac{3}{k}x - \frac{4}{k} \cr & {m_2} = - \frac{3}{k} \cr} $$
Relation between slope of perpendicular lines
$$\eqalign{ & {m_1}{m_2} = - 1 \cr & \Rightarrow \left( {\frac{2}{3}} \right) \times \left( { - \frac{3}{k}} \right) = 1 \cr & \Rightarrow k = 2 \cr} $$