Question

which rule describes the transformation from triangle abc to abc?
image of coordinate grid with triangles abc (green) and abc (blue)
options:
$(x, y) \mapsto (x - 3, y + 4)$
$(x, y) \mapsto (x + 4, y - 3)$
$(x, y) \mapsto (x - 4, y + 3)$
$(x, y) \mapsto (x + 3, y - 4)$

Explanation:

Step1: Identify coordinates of a point

Take point \( B \) with coordinates \( (3, 1) \) and its image \( B' \) with coordinates \( (0, 5) \) (wait, no, looking at the graph, \( B \) is at \( (3, 1) \)? Wait, no, the green triangle: \( B \) is at \( (3, 1) \)? Wait, no, the blue triangle \( B' \) is at \( (0, 5) \)? Wait, no, let's check again. Wait, the green triangle: \( A \) is at \( (2, -4) \), \( C \) is at \( (4, -4) \), \( B \) is at \( (3, 1) \). The blue triangle: \( A' \) is at \( (-1, 0) \), \( C' \) is at \( (1, 0) \), \( B' \) is at \( (0, 5) \). Wait, let's take point \( A \): \( A(2, -4) \) to \( A'(-1, 0) \). Let's calculate the change in \( x \): \( -1 - 2 = -3 \)? No, wait \( -1 - 2 = -3 \)? Wait, no, \( 2 - 4 = -2 \)? Wait, no, maybe I misread. Wait, \( A \) is at \( (2, -4) \), \( A' \) is at \( (-1, 0) \). So \( x \)-change: \( -1 - 2 = -3 \)? No, \( 2 - 4 = -2 \)? Wait, no, let's check the x-coordinate: \( 2 \) to \( -1 \): \( -1 - 2 = -3 \)? No, \( 2 - 4 = -2 \)? Wait, maybe I made a mistake. Wait, the green triangle: \( A \) is at \( (2, -4) \), \( C \) at \( (4, -4) \), \( B \) at \( (3, 1) \). The blue triangle: \( A' \) at \( (-1, 0) \), \( C' \) at \( (1, 0) \), \( B' \) at \( (0, 5) \). So for point \( A \): \( x \)-coordinate: \( -1 - 2 = -3 \)? No, \( 2 - 4 = -2 \)? Wait, no, \( 2 - 4 = -2 \)? Wait, \( 2 - 4 = -2 \), but \( -1 - 2 = -3 \). Wait, maybe take point \( B \): \( B(3, 1) \) to \( B'(0, 5) \). \( x \)-change: \( 0 - 3 = -3 \)? No, \( 0 - 3 = -3 \), \( y \)-change: \( 5 - 1 = 4 \). So \( (x, y) \to (x - 3, y + 4) \)? Wait, no, wait \( B \) is at \( (3, 1) \), \( B' \) is at \( (0, 5) \). So \( x \): \( 3 - 3 = 0 \), \( y \): \( 1 + 4 = 5 \). Yes! So the transformation is \( (x, y) \to (x - 3, y + 4) \)? Wait, no, wait the options: first option is \( (x, y) \mapsto (x - 3, y + 4) \)? Wait, no, the first option is \( (x, y) \mapsto (x - 3, y + 4) \)? Wait, let's check point \( A \): \( A(2, -4) \). \( x - 3 = 2 - 3 = -1 \), \( y + 4 = -4 + 4 = 0 \). Which matches \( A'(-1, 0) \). Point \( C \): \( C(4, -4) \). \( x - 3 = 4 - 3 = 1 \), \( y + 4 = -4 + 4 = 0 \). Which matches \( C'(1, 0) \). Point \( B \): \( B(3, 1) \). \( x - 3 = 3 - 3 = 0 \), \( y + 4 = 1 + 4 = 5 \). Which matches \( B'(0, 5) \). So the correct rule is \( (x, y) \mapsto (x - 3, y + 4) \). Wait, but let's check the options again. The first option is \( (x, y) \mapsto (x - 3, y + 4) \)? Wait, the first option is written as \( (x, y) \mapsto (x - 3, y + 4) \)? Wait, the user's options: first option is \( (x, y) \mapsto (x - 3, y + 4) \), second \( (x + 4, y - 3) \), third \( (x - 4, y + 3) \), fourth \( (x + 3, y - 4) \). So based on the coordinates, the correct transformation is \( (x, y) \mapsto (x - 3, y + 4) \).

Step2: Verify with other points

Check point \( A(2, -4) \): \( x - 3 = -1 \), \( y + 4 = 0 \) (matches \( A'(-1, 0) \)). Point \( C(4, -4) \): \( x - 3 = 1 \), \( y + 4 = 0 \) (matches \( C'(1, 0) \)). Point \( B(3, 1) \): \( x - 3 = 0 \), \( y + 4 = 5 \) (matches \( B'(0, 5) \)). So the rule holds.

Answer:

\((x, y) \mapsto (x - 3, y + 4)\) (the first option)