Q1.
What has remained the same and what has changed with this reflection?
Answer
| Unchanged | Changed |
|---|---|
| All three side lengths: 5, √29, √40 | The sign of every x-coordinate |
| All three angles, so the shape is identical | Position: from Quadrant I to Quadrant II |
| Perimeter, and the area (13 sq units) | Orientation: A→D→M was anticlockwise, A′→D′→M′ is clockwise |
| Every y-coordinate, so every point’s height above the x-axis | Which side of the y-axis each point is on |
| Each point’s distance from the y-axis (its size, not its sign) | Distances from any fixed point that is not on the y-axis |
Why lengths survive but orientation does not: Reflection preserves both shifts in size — the vertical shift exactly, the horizontal shift up to sign — and the distance formula squares them, so every length comes through untouched. A reflection is therefore a congruence: the image triangle can be laid exactly on the original. But it cannot be slid onto it without being turned over, because a mirror reverses the sense in which the vertices are read. That is the one thing a reflection always destroys.
Check it yourself: Going from A (3, 4), the shift to D is (4, −3) and to M is (6, 2). In the image, the shift from A′ to D′ is (−4, −3) and to M′ is (−6, 2) — the horizontal parts have reversed while the vertical parts have not. That mismatch is exactly what flips the triangle over.