9. Given the line $l_{1}: y=4 x$ and the point $P(6,4)$, find a point $Q$ on the line $l_{1}$ such that the line through $P Q$ and the line $l_{1}$ together with the $x$-axis enclose a triangle of minimum area in the first quadrant.
Given $Q(t, 4t)$, then $k_{PQ}=\frac{4t-4}{t-6}$
$\Rightarrow PQ: y=\frac{4t-4}{t-6}(x-6)+4$,
Let $y=0 \Rightarrow A\left(\frac{5t}{t-1}, 0\right)$. Thus $t>1$, and
$S_{\triangle O AQ}=\frac{1}{2} \cdot \frac{5t}{t-1} \cdot 4t=\frac{10(t^2-1)+10}{t-1}$
$=10\left(t-1+\frac{1}{t-1}+2\right) \geqslant 40$, equality holds when $t=2 \Rightarrow Q(2,8)$.
Therefore, when the coordinates of point $Q$ are $(2,8)$, the minimum area of the enclosed triangle is 40.