$25 \cdot 11$ Given the line $y=\frac{3}{4} x+6$, there is another line $l$ parallel to this line, and 4 units away, then a possible equation for $l$ is
(A) $y=\frac{3}{4} x+1$.
(B) $y=\frac{3}{4} x$.
(C) $y=\frac{3}{4} x-\frac{2}{3}$.
(D) $y=\frac{3}{4} x-1$.
(E) $y=\frac{3}{4} x+2$
(14th American High School Mathematics Examination, 1963)
［Solution］As shown in the figure, there are clearly two solutions.
Let the equation of $l_{1}$ be $y=\frac{3}{4} x+b$, where $b$ is the y-intercept, to be determined.
From the figure, it is easy to see that $\triangle A B C \backsim \triangle D A O$, thus
$$
\frac{A B}{A D}=\frac{A C}{D O},
$$

By setting $y=0$, we get $x=-8$; setting $x=0$, we get
$$
y=6 \text {. }
$$

Therefore, $A(0,6), D(-8,0)$. Thus,
$$
A D=\sqrt{6^{2}+8^{2}}=10 .
$$

Hence, $\frac{A B}{10}=\frac{4}{8}$, which gives $A B=5$, and thus
$$
O B=6-5=1 \text {. }
$$

Therefore, the equation of $l_{1}$ is
$$
y=\frac{3}{4} x+1
$$

So the answer is $(A)$.