2019 AMC 10B Problem 5

Reflecting across the line $y = x$ swaps the coordinates: the point $(p, q)$ goes to $(q, p)$. For example, $(1, 2)$ goes to $(2, 1)$. Check each statement with this rule.

1. (A) A point in the first quadrant has both coordinates positive. After swapping, both are still positive. So $A'B'C'$ is in the first quadrant. Always true.
2. (B) A reflection is a mirror image. It does not stretch or shrink anything, so the area stays the same. Always true.
3. (C) Take $A = (p, q)$, so $A' = (q, p)$. The slope of $AA'$ is
$$\frac{p - q}{q - p} = -1.$$
The bottom is not zero because $A$ is not on the line $y = x$, so $p \ne q$. Always true.
4. (D) By step 3, both $AA'$ and $CC'$ have slope $-1$. Always true.
5. (E) Try a concrete example. Let $A = (1, 2)$ and $B = (2, 4)$. Line $AB$ has slope $\frac{4 - 2}{2 - 1} = 2$. The images are $A' = (2, 1)$ and $B' = (4, 2)$, and line $A'B'$ has slope $\frac{2 - 1}{4 - 2} = \frac12$. Perpendicular lines have slopes that multiply to $-1$, but $2 \cdot \frac12 = 1$. So these lines are not perpendicular.

Statement (E) is the one that is not always true.

The answer is $\boxed{\textbf{(E)}\ \text{Lines } AB \text{ and } A'B' \text{ are perpendicular}}$.

Swapping $x$ and $y$ trades "rise" and "run," so a slope $m$ becomes $\frac1m$. The product $m \cdot \frac1m$ is $1$, not $-1$, so a line and its mirror image are usually not perpendicular. For a "not always true" question, one concrete example that fails is all you need.