$25 \cdot 6$ On the coordinate plane, lines $l$ and $k$ are symmetric with respect to the line $y=x$. If the equation of line $l$ is $y=a x+b$, where $a \neq 0, \quad b \neq 0$, then the equation of line $k$ is $y$ equals
(A) $\frac{1}{a} x+b$.
(B) $-\frac{1}{a} x+b$.
(C) $-\frac{1}{a} x-\frac{b}{a}$.
(D) $\frac{1}{a} x+\frac{b}{a}$.
(E) $\frac{1}{a} x-\frac{b}{a}$.
(32nd American High School Mathematics Examination, 1981)
[Solution] The point $(p, q)$ on the line $l$ has its symmetric point $(q, p)$ with respect to the line $y=x$ necessarily on the line $k$.
Therefore, any point on $k$ must satisfy $\quad x=a y+b$, i.e., $y=\frac{x}{a}-\frac{b}{a}$. Hence, the correct choice is $(E)$.