Question

complete the table to show the coordinates of the vertices of △abc.
△abc
a(-2,3)
b(0,4)
c(1,1)
△abc
?
a(1,3)
a(4,1)
a(3,4)

Explanation:

Step1: Analyze the transformation

Looking at the coordinates of \( B(0,4) \) and its image \( A'(1,3) \)? Wait, no, looking at the graph, we can see the translation. Let's check the horizontal shift. For point \( A(-2,3) \), looking at the graph, \( A' \) should be at \( (1,3) \)? Wait, no, let's check the x - coordinate change. From \( A(-2,3) \) to \( A' \): let's see the horizontal shift. The original \( A \) is at \( x=-2 \), and the new \( A' \) in the graph is at \( x = 1 \). The change in x - coordinate is \( 1-(-2)=3 \)? Wait, no, looking at the table, for \( B(0,4) \), the image is \( A'(1,3) \)? No, the table has a typo? Wait, no, looking at the graph, \( A \) is at \( (-2,3) \), \( A' \) is at \( (1,3) \)? Wait, no, the graph shows \( A \) at \( (-2,3) \), \( A' \) at \( (1,3) \)? Wait, the x - coordinate changes from - 2 to 1, which is a shift of \( 1-(-2)=3 \)? Wait, no, let's check the horizontal distance. From \( A(-2,3) \) to \( A' \): looking at the graph, the x - coordinate of \( A \) is - 2, and \( A' \) is at x = 1. So the translation vector is (3,0)? Wait, no, let's check \( B(0,4) \). If we shift \( B(0,4) \) by 3 units to the right, we get \( (0 + 3,4)=(3,4) \)? But the table has \( A'(1,3) \) for \( B(0,4) \), which is wrong. Wait, no, the table is mislabeled. Looking at the graph, \( A \) is at \( (-2,3) \), \( A' \) is at \( (1,3) \) (x - coordinate: - 2+3 = 1, y - coordinate: 3). \( B(0,4) \) should be \( B'(0 + 3,4)=(3,4) \)? Wait, no, the graph shows \( B \) at \( (0,4) \), \( B' \) at \( (3,4) \)? Wait, the graph has \( A' \) at \( (1,3) \)? No, the graph's \( A' \) is at (1,3)? Wait, no, the graph: \( A \) is at (-2,3), \( A' \) is at (1,3) (x: - 2 + 3=1, y:3). So for \( A(-2,3) \), the image \( A' \) is \( (-2 + 3,3)=(1,3) \).

Step2: Apply the translation to \( A(-2,3) \)

The translation vector is (3,0) (since x - coordinate increases by 3, y - coordinate remains the same). So for \( A(-2,3) \), \( A'=(-2 + 3,3)=(1,3) \). Wait, but the table has \( A'(1,3) \) for \( B(0,4) \), which is a mislabel. Correctly, for \( A(-2,3) \), the image \( A' \) is \( (1,3) \).

Answer:

\( A'(1,3) \)