Algebra + Geometry = Co - ordinate Geometry

Since region $OPQRST$ has area $48$ we need the $2$ straight lines to divide the region into $3$ subregions each of area $16$ .

Note first that $\Delta OPQ$ has area $18$ , so we know that one line will need to intersect $PQ$ at some point $A(c,6)$ where $0 \lt c \lt 6$ . We then need to find $c$ such that $\Delta OPA$ has area $16$ , which will be the case when

$(\frac{1}{2})*6c = 16 \Longrightarrow c = \frac{16}{3}$ .

This gives a slope of this first line of $\dfrac{6}{\frac{16}{3}} = \dfrac{9}{8}$ .

To find the second line, first note that region $ORST$ has area $18$ , so we will need the second line to intersect segment $RS$ at some point $B(d,2)$ where $6 \lt d \lt 12$ . We then need to find $d$ such that the region $OBST$ has area $16$ , which will be the case when

$(\frac{1}{2})*2d + (12 - d)*2 = 16 \Longrightarrow 24 - d = 16 \Longrightarrow d = 8$ .

Thus the slope of the second line is $\frac{2}{8} = \frac{1}{4}$ , and so the sum of the slopes is

$\frac{9}{8} + \frac{2}{8} = \frac{11}{8}$ .

We thus have $a + b = 11 + 8 = \boxed{19}$ .