Note that the slope of $ax+y=1$ is $-a$ and the slope of $x+by=1$ is $\dfrac{-1}{b}$.
A is false. According to the figure, the slope of $ax+y=1$ is positive. Hence, we have
$\begin{array}{rcl}
-a & > & 0 \\
a & < & 0
\end{array}$
B is false. According to the figure, the slope of $x+by=1$ is positive. Hence, we have
$\begin{array}{rcl}
\dfrac{-1}{b} & > & 0 \\
b & < & 0
\end{array}$
C is false. Consider the intersection point of the two line. It lies on the 3rd quadrant, the $x$ and $y$ coordinates are negative.
$\left\{ \begin{array}{ll}
ax+y=1 & \ldots \unicode{x2460} \\
x+by=1 & \ldots \unicode{x2461}
\end{array}\right.$
From $\unicode{x2460}$, we have
$\begin{array}{rcl}
ax+y & = & 1 \\
y & = & 1- ax ~\ldots \unicode{x2462}
\end{array}$
Sub. $\unicode{x2462}$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
x + b( 1-ax) & = & 1 \\
x + b – abx & = & 1 \\
x (1-ab) & = & 1-b \\
x & = & \dfrac{1-b}{1-ab}
\end{array}$
Since the $x$ coordinate of the intersection is negative, then $\dfrac{1-b}{1-ab} < 0$. By the above argument, $b<0$. Therefore, we have
$\begin{array}{rcl}
1-ab & < & 0 \\
ab & > & 1
\end{array}$
Hence, D is true.
