Question

The image of trapezoid ABCD has coordinates A′(–2, –5), B′(–1, –2), C′(2, –2), and D′(3, –5). It was translated by the rule T–1, –3(x, y). Which diagram shows the pre-image?

On a coordinate plane, a trapezoid has points A (negative 1, negative 2), B (0, 1), C (3, 1), D (4, negative 2).

On a coordinate plane, a trapezoid has points A (negative 3, negative 2), B (negative 2, 1), C (1, 1), D (2, negative 2).

On a coordinate plane, a trapezoid has points A (negative 4, negative 5), B (negative 3, negative 2), C (0, negative 2), D (1, negative 5).

On a coordinate plane, a trapezoid has points A (0, negative 2), B (1, 1), C (4, 1), D (5, negative 2).

Answer 1

Translation T_{-1,-3} maps (x,y) to (x-1, y-3), so to get the pre-image add (1,3) to each image point.

A: (-2,-5) → (-1,-2)
B: (-1,-2) → (0,1)
C: (2,-2) → (3,1)
D: (3,-5) → (4,-2)

So the first diagram (A(−1,−2), B(0,1), C(3,1), D(4,−2)) is the pre-image.