Matemática Elementar

Problema 1

Triangle $ABC$ has vertices $A(1, 2)$ , $B(4, 2)$ , and $C(4, 6)$ . Apply a reflection over the $y$ -axis to obtain triangle $A'B'C'$ . List the coordinates of $A'$ , $B'$ , and $C'$ , and explain why the two triangles are congruent.

Problemas

1. Triangle $ABC$ has vertices $A(1, 2)$ , $B(4, 2)$ , and $C(4, 6)$ . Apply a reflection over the $y$ -axis to obtain triangle $A'B'C'$ . List the coordinates of $A'$ , $B'$ , and $C'$ , and explain why the two triangles are congruent.

2. Point $P(3, -1)$ is rotated $90°$ counterclockwise about the origin. What are the coordinates of the image $P'$ ? Then $P'$ is translated by the vector $\langle 2, 5 \rangle$ . What are the final coordinates $P''$ ?

3. Quadrilateral $DEFG$ has vertices $D(0, 0)$ , $E(3, 0)$ , $F(3, 2)$ , and $G(0, 2)$ . A classmate claims that rotating the quadrilateral $180°$ about the origin and then translating it by $\langle 3, 2 \rangle$ produces an image that coincides exactly with the original. Is the classmate correct? Justify your answer with coordinates.

4. Two triangles are shown in the coordinate plane. Triangle $RST$ has vertices $R(-5, 1)$ , $S(-2, 1)$ , and $T(-2, 4)$ . Triangle $R'S'T'$ has vertices $R'(1, -5)$ , $S'(1, -2)$ , and $T'(4, -2)$ . Describe a single rigid motion that maps $RST$ onto $R'S'T'$ . Is $RST \cong R'S'T'$ ? Explain.

5. Segment $\overline{AB}$ has endpoints $A(2, 3)$ and $B(6, 7)$ . Segment $\overline{CD}$ has endpoints $C(-3, 2)$ and $D(-7, 6)$ . Show that $\overline{AB} \cong \overline{CD}$ by computing both lengths, and then describe a sequence of rigid motions (reflection and translation) that maps $\overline{AB}$ onto $\overline{CD}$ .

Gabarito — Congruence and Transformations

1. A reflection over the $y$ -axis maps $(x, y) \to (-x, y)$ .

$A(1, 2) \to A'(-1, 2)$ $B(4, 2) \to B'(-4, 2)$ $C(4, 6) \to C'(-4, 6)$

The two triangles are congruent because a reflection is a rigid motion — it preserves distances and angle measures. Every side of $\triangle ABC$ has the same length as the corresponding side of $\triangle A'B'C'$ , and every angle is preserved. Since one triangle is the image of the other under a rigid motion, $\triangle ABC \cong \triangle A'B'C'$ .

2. A $90°$ counterclockwise rotation about the origin maps $(x, y) \to (-y, x)$ .

$P(3, -1) \to P'(-(-1), 3) = P'(1, 3)$

Now translate $P'(1, 3)$ by $\langle 2, 5 \rangle$ :

$P''(1 + 2,\; 3 + 5) = P''(3, 8)$

3. A $180°$ rotation about the origin maps $(x, y) \to (-x, -y)$ .

$D(0, 0) \to (0, 0), \quad E(3, 0) \to (-3, 0), \quad F(3, 2) \to (-3, -2), \quad G(0, 2) \to (0, -2)$

Now translate each image point by $\langle 3, 2 \rangle$ (add $3$ to $x$ , add $2$ to $y$ ):

$D'(0+3, 0+2) = (3, 2)$ $E'(-3+3, 0+2) = (0, 2)$ $F'(-3+3, -2+2) = (0, 0)$ $G'(0+3, -2+2) = (3, 0)$

The image vertices are $(3,2)$ , $(0,2)$ , $(0,0)$ , $(3,0)$ , which are exactly the set $\{D, E, F, G\} = \{(0,0), (3,0), (3,2), (0,2)\}$ . The images coincide with the original vertices (in a different labeling order: $D' = F$ , $E' = G$ , $F' = D$ , $G' = E$ ).

Yes, the classmate is correct. The composed rigid motion maps the rectangle onto itself.

4. Consider the mapping $(x, y) \to (y, x)$ , which is a reflection over the line $y = x$ .

$R(-5, 1) \to (1, -5) = R'$ $S(-2, 1) \to (1, -2) = S'$ $T(-2, 4) \to (4, -2) = T'$

Every vertex of $\triangle RST$ maps to the corresponding vertex of $\triangle R'S'T'$ under reflection over $y = x$ .

Since a reflection is a rigid motion (preserving distances and angles), $\triangle RST \cong \triangle R'S'T'$ .

5. Compute both lengths using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .

$AB = \sqrt{(6-2)^2 + (7-3)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$

$CD = \sqrt{(-7-(-3))^2 + (6-2)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$

Since $AB = CD = 4\sqrt{2}$ , the segments are congruent: $\overline{AB} \cong \overline{CD}$ .

Rigid motion sequence:

Step 1 — Reflect over the $y$ -axis: $(x, y) \to (-x, y)$ .

$A(2, 3) \to (-2, 3), \quad B(6, 7) \to (-6, 7)$

Step 2 — Translate by $\langle -1, -1 \rangle$ :

$(-2, 3) \to (-3, 2) = C, \quad (-6, 7) \to (-7, 6) = D$

The composition of a reflection followed by a translation maps $\overline{AB}$ onto $\overline{CD}$ .