Explanation:

Drop a perpendicular from D(-4, 6) to x-axis (as shown in the figure)


AC = $x$
CO = 4
CD = OE = 6
BE = $y$

△ACD ~ △AOE
i.e. $\frac{x}{(x+4)} = \frac{6}{(6+y)} = \sqrt{\frac{Ar. (△ACD)}{ Ar. (△AOE)}} = \sqrt{\frac{3x}{54}}$

Solve, $\frac{x}{(x+4)} = \sqrt{\frac{3x}{54}} = \sqrt{\frac{x}{18}}$

Squaring both sides;
$\frac{x^2}{(x+4)^2} = \frac{x}{18}$

$\frac{x}{(x+4)^2} = \frac{1}{18}$

$18x = x^2 + 8x + 16$
$x^2 - 10x + 16 = 0$
$x^2 - 8x - 2x + 16 = 0$
$x(x-8) - 2(x-8) = 0$
$(x-2)(x-8) = 0$
$x = 2$ or $8$

Taking $x=8$, and solve for $y$;
$\frac{8}{(8+4)} = \frac{6}{(6+y)}$
$48+8y = 72 + 12y$
$4y = 36$
$y=9$

Therefore, A is (-12, 0) and B(0, 9)

Apply slope formulae;
$m = \frac{(9-0)}{(0+12)} = \frac{9}{12} = \frac{3}{4}$

Hence, option B

[NOTE: if you take $x=2$, you get $m=3$ - which is not in the option choices]