1. In the Cartesian coordinate system $x O y$, the line $l: y=$ $-\frac{4}{3} x+4$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively. If the line $l$ is rotated counterclockwise by $90^{\circ}$ around point $O$, it intersects the $y$-axis at point $A^{\prime}$ and the line $l$ at point $C$. Then the area of $\triangle A^{\prime} B C$ is ( ).
(A) $\frac{3}{25}$
(B) $\frac{6}{25}$
(C) $\frac{12}{25}$
(D) $\frac{24}{25}$
$-1 . B$.
As shown in Figure 3, let the line $l$ rotate around point $O$ and intersect the $x$-axis at $B'$. It is easy to know that $O A=3$, $O B=4$, so $O A'=3$, $O B'=4$. Therefore, $A' B=1$.
By the Pythagorean theorem, we have
$$
A B=5 \text {. }
$$

By the properties of rotation, we know $\angle A' C B=90^{\circ}$. Hence, $\triangle A' C B \backsim \triangle A O B$, and
$$
\frac{S_{\triangle A' C B}}{S_{\triangle A O B}}=\left(\frac{A' B}{A B}\right)^{2}=\left(\frac{1}{5}\right)^{2}=\frac{1}{25} \text {. }
$$

Since $S_{\triangle A O B}=\frac{1}{2} O A \cdot O B=\frac{1}{2} \times 3 \times 4=6$, then
$$
S_{\triangle A' C B}=\frac{1}{25} S_{\triangle A O B}=\frac{6}{25} .
$$