1. If the point $A(3,5)$ is symmetric to a point on the $x$-axis with respect to the line $l: y=k x$, then $k$ is ( ).
(A) $\frac{-1 \pm \sqrt{5}}{2}$
(B) $\pm \sqrt{3}$
(C) $\frac{-1 \pm \sqrt{30}}{4}$
(D) $\frac{-3 \pm \sqrt{34}}{5}$
-、1.D.
Let point $A(3,5)$ have its symmetric point with respect to the line $l: y=k x$ as $P(a, 0)$. Then the midpoint $M\left(\frac{3+a}{2}, \frac{5}{2}\right)$ of $A P$ must lie on the line $l$, and $A P \perp l$, i.e., $\frac{5}{2}=k \cdot \frac{3+a}{2}$, and $k \cdot \frac{5-0}{3-a}=-1$.
Eliminating $a$ and rearranging gives $5 k^{2}+6 k-5=0$.
Solving yields $k=\frac{-3 \pm \sqrt{34}}{5}$.