Let $$A(-1, 1)$$, $$B(3, 4)$$ and $$C(2, 0)$$ be given three points. A line $$y = mx$$, $$m > 0$$, intersects lines $$AC$$ and $$BC$$ at point $$P$$ and $$Q$$ respectively. Let $$A_1$$ and $$A_2$$ be the areas of $$\triangle ABC$$ and $$\triangle PQC$$ respectively, such that $$A_1 = 3A_2$$, then the value of $$m$$ is equal to:

We are given $$A(-1, 1)$$, $$B(3, 4)$$, $$C(2, 0)$$ and a line $$y = mx$$ with $$m > 0$$ that intersects line $$AC$$ at $$P$$ and line $$BC$$ at $$Q$$. We need $$A_1 = 3A_2$$ where $$A_1$$ and $$A_2$$ are areas of $$\triangle ABC$$ and $$\triangle PQC$$.

The equation of line $$AC$$ (passing through $$A(-1,1)$$ and $$C(2,0)$$) is $$y = \frac{2-x}{3}$$. Setting $$y = mx$$: $$mx = \frac{2-x}{3}$$, giving $$x = \frac{2}{3m+1}$$. So $$P = \left(\frac{2}{3m+1}, \frac{2m}{3m+1}\right)$$.

The equation of line $$BC$$ (passing through $$B(3,4)$$ and $$C(2,0)$$) is $$y = 4(x-2)$$. Setting $$y = mx$$: $$mx = 4x - 8$$, giving $$x = \frac{8}{4-m}$$. So $$Q = \left(\frac{8}{4-m}, \frac{8m}{4-m}\right)$$.

Since $$P$$ lies on $$AC$$ and $$Q$$ lies on $$BC$$, triangles $$PQC$$ and $$ABC$$ share the angle at $$C$$, so $$\frac{A_2}{A_1} = \frac{CP}{CA} \cdot \frac{CQ}{CB}$$.

Computing $$\frac{CP}{CA}$$: $$CP^2 = \frac{36m^2 + 4m^2}{(3m+1)^2} = \frac{40m^2}{(3m+1)^2}$$. Since $$CA^2 = 10$$, we get $$\frac{CP}{CA} = \frac{2m}{3m+1}$$.

Computing $$\frac{CQ}{CB}$$: $$CQ^2 = \frac{4m^2 + 64m^2}{(4-m)^2} = \frac{68m^2}{(4-m)^2}$$. Since $$CB^2 = 17$$, we get $$\frac{CQ}{CB} = \frac{2m}{4-m}$$.

Setting $$\frac{A_2}{A_1} = \frac{1}{3}$$: $$\frac{4m^2}{(3m+1)(4-m)} = \frac{1}{3}$$. This gives $$12m^2 = -3m^2 + 11m + 4$$, so $$15m^2 - 11m - 4 = 0$$.

Using the quadratic formula: $$m = \frac{11 \pm \sqrt{121 + 240}}{30} = \frac{11 \pm 19}{30}$$. This gives $$m = 1$$ or $$m = -\frac{8}{30}$$. Since $$m > 0$$, we have $$m = 1$$.