imc

Let's analyze all of the options separately. $\text{(A)}$: Clearly $\text{(A)}$ 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. $\text{(B)}$: The triangles have the same area, since $△ABC$ and $△A^{\prime }{B}^{\prime }{C}^{\prime }$ are the same triangle (congruent). More formally, we can say that area is invariant under reflection. $\text{(C)}$: If point $A$ has coordinates $(p,q)$, then $A^{\prime }$ will have coordinates $(q,p)$. The gradient is thus $\frac{p-q}{q-p}=-1$, so this is true. (We know $p\ne q$ 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). $\text{(D)}$: Repeating the argument for $\text{(C)}$, we see that both lines have slope $-1$, so this is also true. $\text{(E)}$: 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 $A^{\prime }$ and $B^{\prime }$ will, respectively, have coordinates $(q,p)$ and $(s,r)$. The product of the gradients of $AB$ and $A^{\prime }{B}^{\prime }$ is $\frac{s-q}{r-p}\cdot \frac{r-p}{s-q}=1\ne -1$, so in fact these lines are never perpendicular to each other (using the "negative reciprocal" condition for perpendicularity). Thus the answer is $\boxed{\text{Lines }AB\text{ and }{A}^{\prime }{B}^{\prime }\text{ are perpendicular to each other.}}$.