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2019 AMC 10B Problem 5

2019 AMC 10B Problem 5AMC 10 Step by Step

Name: Date: October 28, 2026

Coordinate Geometry · difficulty 1 of 5 · about 2 min · position P1-10

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?

A)Triangle $A'B'C'$ lies in the first quadrant.
B)Triangles $ABC$ and $A'B'C'$ have the same area.
C)The slope of line $AA'$ is $-1$ .
D)The slopes of lines $AA'$ and $CC'$ are the same.
E)Lines $AB$ and $A'B'$ are perpendicular to each other.

Answer

Key insight

Reflecting over y = x swaps the two coordinates of each point. So a line of slope 2 becomes a line of slope 1/2, and those are not perpendicular.

Solution

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

Why this works

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.

The trap

Thinking a line and its mirror image across y = x must be perpendicular. They are only perpendicular in special cases.

Concise solution

Reflection across $y = x$ sends $(p, q)$ to $(q, p)$ .

(A) Swapping two positive coordinates gives two positive coordinates, so $A'B'C'$ stays in the first quadrant. Always true.
(B) Reflections are rigid motions, so area is preserved. Always true.
(C) For $A = (p, q)$ with $p \ne q$ , the slope of $AA'$ is $\frac{p - q}{q - p} = -1$ . Always true.
(D) By (C), every such segment has slope $-1$ . Always true.
(E) Try $A = (1, 2)$ , $B = (2, 4)$ : line $AB$ has slope $2$ . Then $A' = (2, 1)$ , $B' = (4, 2)$ , and $A'B'$ has slope $\frac12$ . The product of the slopes is $1$ , not $-1$ , so the lines are not perpendicular.

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

Review in the AoPS books

Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Geometry, Chapter 16: The More Things Change. . . (§16.1, §16.2, §16.3, §16.4)
Introduction to Algebra, Chapter 8: Graphing Lines

Techniques

Use the answer choices (mod checks, size, form) to eliminate or select · Test small/specific values or special cases to find or verify the answer