Question: The graph shows triangles DEF and OPQ.
Is DEF congruent to OPQ? Justify your answer.
Yes, because a reflection across the y-axis followed by a translation left 2 units and down 1 unit maps DEF onto OPQ.
Yes, because a rotation 90° counterclockwise around the origin followed by a reflection across the x-axis maps DEF onto OPQ.
No, because EF and PQ do not have the same length.
No, because \angle D and \angle O do not have the same measure.
1. **State the problem:** Determine if triangles DEF and OPQ are congruent based on the given transformations and side/angle comparisons.
2. **Recall congruence criteria:** Triangles are congruent if one can be mapped onto the other by rigid motions (translations, rotations, reflections) preserving side lengths and angles.
3. **Check the given transformations:**
- Reflection across the y-axis followed by translation left 2 units and down 1 unit.
- Rotation 90\degree counterclockwise around the origin followed by reflection across the x-axis.
4. **Calculate side lengths of DEF:**
- $D(-2,3), E(6,-2), F(4,4)$
- $DE = \sqrt{(6 - (-2))^2 + (-2 - 3)^2} = \sqrt{8^2 + (-5)^2} = \sqrt{64 + 25} = \sqrt{89}$
- $EF = \sqrt{(4 - 6)^2 + (4 - (-2))^2} = \sqrt{(-2)^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40}$
- $FD = \sqrt{(4 - (-2))^2 + (4 - 3)^2} = \sqrt{6^2 + 1^2} = \sqrt{36 + 1} = \sqrt{37}$
5. **Calculate side lengths of OPQ:**
- $O(-3,2), P(2,-6), Q(-4,-4)$
- $OP = \sqrt{(2 - (-3))^2 + (-6 - 2)^2} = \sqrt{5^2 + (-8)^2} = \sqrt{25 + 64} = \sqrt{89}$
- $PQ = \sqrt{(-4 - 2)^2 + (-4 - (-6))^2} = \sqrt{(-6)^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40}$
- $QO = \sqrt{(-4 - (-3))^2 + (-4 - 2)^2} = \sqrt{(-1)^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37}$
6. **Compare side lengths:**
- $DE = OP = \sqrt{89}$
- $EF = PQ = \sqrt{40}$
- $FD = QO = \sqrt{37}$
7. **Check angles:** Since side lengths correspond, angles must also correspond by the Side-Side-Side (SSS) congruence criterion.
8. **Check transformations:** The rotation 90\degree counterclockwise around the origin followed by reflection across the x-axis maps DEF onto OPQ, confirming congruence.
**Final answer:**
Yes, because a rotation 90\degree counterclockwise around the origin followed by a reflection across the x-axis maps DEF onto OPQ, confirming the triangles are congruent by SSS congruence.