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]