Question

Which algebraic rule describes the translation of ΔDEF to ΔD'E'F'?
An image displaying two triangles DEF and D'E'F' in a graph with vertices. DEF with vertices D of -1 comma 5, E of 3 comma 3 and F of -1 comma 3 and D'E'F' with vertices D' of 3 comma 2, E' of 7 comma 0 and F' of 3 comma0

A.
(x, y) →(x + 3, y − 5)
B.
(x, y) →(x − 4, y + 3)
C.
(x, y) →(x + 4, y − 3)
D.
(x, y) →(x – 3, y + 4)

Answer 1

To find the translation from triangle ΔDEF to triangle ΔD'E'F', we need to determine how the coordinates of each vertex of triangle DEF change to reach the corresponding vertex in triangle D'E'F'.

The coordinates of the vertices of triangle DEF are:
- D(-1, 5)
- E(3, 3)
- F(-1, 3)

The coordinates of the corresponding vertices of triangle D'E'F' are:
- D'(3, 2)
- E'(7, 0)
- F'(3, 0)

Let's calculate the translation for each vertex:

1. For point D to D':
- From D(-1, 5) to D'(3, 2):
- x-coordinate: \(3 - (-1) = 4\) (increase by 4)
- y-coordinate: \(2 - 5 = -3\) (decrease by 3)

Thus, the translation from D to D' is:
\[
(x, y) \to (x + 4, y - 3)
\]

2. For point E to E':
- From E(3, 3) to E'(7, 0):
- x-coordinate: \(7 - 3 = 4\) (increase by 4)
- y-coordinate: \(0 - 3 = -3\) (decrease by 3)

Thus, the translation from E to E' confirms:
\[
(x, y) \to (x + 4, y - 3)
\]

3. For point F to F':
- From F(-1, 3) to F'(3, 0):
- x-coordinate: \(3 - (-1) = 4\) (increase by 4)
- y-coordinate: \(0 - 3 = -3\) (decrease by 3)

Thus, the translation from F to F' is also:
\[
(x, y) \to (x + 4, y - 3)
\]

Since all vertices undergo the same transformation, the rule that describes the translation of ΔDEF to ΔD'E'F' is:

**C. (x, y) → (x + 4, y - 3)**.