1. Given $m>2$, line $l_{1}: y=\frac{m-2}{m} x+2$, line $l_{2}: y=-x+2 m$ and the $y$-axis form a triangle with an area of 30. Then the value of $m$ is ( ).
(A)6
(B) 12
(C) $\frac{1+\sqrt{61}}{2}$
(D) $1+\sqrt{61}$
-,1.A.
Notice that, line $l_{1}$ intersects the $y$-axis at point $(0,2)$, line $l_{2}$ intersects the $y$-axis at point $(0,2 m)$, and the intersection point of lines $l_{1}$ and $l_{2}$ is $(m, m)$.
Then $\frac{1}{2} m(2 m-2)=30 \Rightarrow m=6$.