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2019 AMC 10B Problem 5

2019 AMC 10B Problem 5AMC 10 Step by Step

Name: Date: September 13, 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 coordinates, so a segment of slope m becomes one of slope 1/m; the product is 1, not -1, so they are not perpendicular.

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

Why this works

Reflecting across $y = x$ turns a slope $m$ into $\frac1m$ (rise and run trade places). Two lines are perpendicular when their slopes multiply to $-1$ , but $m \cdot \frac1m = 1$ , so mirror images across $y = x$ are perpendicular only in the special horizontal/vertical case. For "not always true" questions, one concrete counterexample settles it.

The trap

Believing mirror-image lines across y = x are always perpendicular, or thinking the reflected triangle can leave the first quadrant.

Common mistakes

Believing mirror-image lines across y = x are always perpendicular, or thinking the reflected triangle can leave the first quadrant.
Doubting (C) or (D) because the slope looks like it should depend on the point; the segment from a point to its mirror image is always perpendicular to the mirror line $y = x$ .

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