7. Through point $M(2,1)$, draw a line $l$ that intersects the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ at points $A, B$. If point $M$ is exactly the trisection point of segment $A B$, then the slope of line $l$ is . $\qquad$
7. $\frac{-4 \pm \sqrt{7}}{6}$

Analysis: Given that the line $\boldsymbol{Z}$ passes through the point $M(2,1)$, its parametric equation is: $\left\{\begin{array}{l}x=2+t \cos \alpha \\ y=1+t \sin \alpha\end{array} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . . . . . . .\right.$.
Thus, the slope of the line $l$ is $\tan \alpha$. Substituting $(*)$ into the equation of the ellipse, we get:
$(2+t \cos \alpha)^{2}+4(1+t \sin \alpha)^{2}-16=0$, simplifying to:
$\left(1+3 \sin ^{2} \alpha\right) \cdot t^{2}+(4 \cos \alpha+8 \sin \alpha) \cdot t-8=0$. By Vieta's formulas:
$t_{1}+t_{2}=-\frac{4 \cos \alpha+8 \sin \alpha}{1+3 \sin ^{2} \alpha}, t_{1} \cdot t_{2}=-\frac{8}{1+3 \sin ^{2} \alpha}$. Since $M$ is exactly the point that trisects the segment $A B$, we have
$\frac{t_{1}}{t_{2}}=-2$ or $-\frac{1}{2}$, which means $\frac{\left(t_{1}+t_{2}\right)^{2}}{t_{1} \cdot t_{2}}=-\frac{(4 \cos \alpha+8 \sin \alpha)^{2}}{8\left(1+3 \sin ^{2} \alpha\right)}=-\frac{1}{2}$, hence
$(2 \cos \alpha+4 \sin \alpha)^{2}=\cos ^{2} \alpha+4 \sin ^{2} \alpha \Leftrightarrow 3 \cos ^{2} \alpha+12 \sin ^{2} \alpha+16 \sin \alpha \cos \alpha=0$
$12 \tan ^{2} \alpha+16 \tan \alpha+3=0 \Leftrightarrow \tan \alpha=\frac{-4 \pm \sqrt{7}}{6}$