1. Let $x, y \in R$, set $A=\left\{(x, y) \left\lvert\, x^{2}-3 x y+4 y^{2}=\frac{7}{2}\right.\right\} ; B=\{(x, y) \mid k x+y=2\}(k>0)$, when $A \mathrm{I} B$ has only one element, the value of $\mathrm{k}$ is $\qquad$
1. $\frac{1}{4}$

Analysis: Solving the system of equations $\left\{\begin{array}{c}x^{2}-3 x y+4 y^{2}=\frac{7}{2} \\ k x+y=2\end{array}\right.$, we get $x^{2}-3 x(2-k x)+4(2-k x)^{2}=\frac{7}{2}$, simplifying to $\left(1+3 k+4 k^{2}\right) x^{2}-(6+16 k) x+\frac{25}{2}=0$. When $A \mid B$ has only one element, $\Delta=0$, i.e., $(6+16 k)^{2}-50\left(1+3 k+4 k^{2}\right)=0 \Leftrightarrow 4 k^{2}+3 k-1=0 \Leftrightarrow k=-1$ or $\frac{1}{4}$. Since $k>0$, thus $k=\frac{1}{4}$.