Triangle $ABC$ lies in the first quadrant. Points $A$, $B$, and $C$ are reflected across the line $y=x$ to points $A'$, $B'$, and $C'$, respectively. Assume that none of the vertices of the triangle lie on the line $y=x$. Which of the following statements is not always true?
$\textbf{(A) }$ Triangle $A'B'C'$ lies in the first quadrant.
$\textbf{(B) }$ Triangles $ABC$ and $A'B'C'$ have the same area.
$\textbf{(C) }$ The slope of line $AA'$ is $-1$.
$\textbf{(D) }$ The slopes of lines $AA'$ and $CC'$ are the same.
$\textbf{(E) }$ Lines $AB$ and $A'B'$ are perpendicular to each other.
Let's analyze all of the options separately. 
$\textbf{(A)}$: Clearly $\textbf{(A)}$ is true, because a point in the first quadrant will have non-negative $x$- and $y$-coordinates, and so its reflection, with the coordinates swapped, will also have non-negative $x$- and $y$-coordinates.
$\textbf{(B)}$: The triangles have the same area, since $\triangle ABC$ and $\triangle A'B'C'$ are the same triangle (congruent). More formally, we can say that area is invariant under reflection.
$\textbf{(C)}$: If point $A$ has coordinates $(p,q)$, then $A'$ will have coordinates $(q,p)$. The gradient is thus $\frac{p-q}{q-p} = -1$, so this is true. (We know $p \neq q$ since the question states that none of the points $A$, $B$, or $C$ lies on the line $y=x$, so there is no risk of division by zero).
$\textbf{(D)}$: Repeating the argument for $\textbf{(C)}$, we see that both lines have slope $-1$, so this is also true.
$\textbf{(E)}$: This is the only one left, presumably the answer. To prove: if point $A$ has coordinates $(p,q)$ and point $B$ has coordinates $(r,s)$, then $A'$ and $B'$ will, respectively, have coordinates $(q,p)$ and $(s,r)$. The product of the gradients of $AB$ and $A'B'$ is $\frac{s-q}{r-p} \cdot \frac{r-p}{s-q} = 1 \neq -1$, so in fact these lines are never perpendicular to each other (using the "negative reciprocal" condition for perpendicularity).
Thus the answer is $\boxed{\textbf{(E)}}$.