Question

question 2 of 10 which of the following steps were applied to δabc obtain δabc? a. shifted 2 units right and 3 units up b. shifted 4 units right and 3 units up

Explanation:

Step1: Identify coordinates of a point

Let's take point \( A \) from \( \triangle ABC \). From the graph, \( A \) has coordinates \( (2, 5) \) (assuming each grid is 1 unit). Point \( A' \) in \( \triangle A'B'C' \) has coordinates \( (4, 8) \)? Wait, no, wait. Wait, looking at the graph, let's recheck. Wait, \( A \) is at \( (2, 5) \)? Wait, no, maybe \( A \) is at \( (2, 5) \)? Wait, no, let's see the grid. The x-axis: from left, the first grid after 0 is 1, 2, etc. Wait, \( A \) is at \( (2, 5) \)? Wait, \( A' \) is at \( (4, 8) \)? No, wait, maybe \( A \) is at \( (2, 5) \), \( A' \) is at \( (4, 8) \)? Wait, no, let's check the horizontal (x) and vertical (y) shifts. Let's take point \( B \): \( B \) is at \( (3, 10) \)? Wait, no, the graph: \( B \) is at \( (3, 10) \)? Wait, \( B' \) is at \( (5, 13) \)? No, the y-axis: the grid lines, \( B \) is at \( (3, 10) \), \( B' \) is at \( (5, 13) \)? No, that can't be. Wait, maybe I misread. Wait, the original triangle \( \triangle ABC \): let's find coordinates of \( A \), \( B \), \( C \). Let's assume each square is 1 unit. So \( A \) is at \( (2, 5) \), \( B \) at \( (3, 10) \), \( C \) at \( (6, 5) \)? Wait, no, \( A' \) is at \( (4, 8) \), \( B' \) at \( (5, 13) \)? No, that's not matching. Wait, maybe \( A \) is at \( (2, 5) \), \( A' \) at \( (4, 8) \): the x-shift is \( 4 - 2 = 2 \)? No, wait, no, let's look again. Wait, the correct way: take a point, say \( A \): original \( A \) is at \( (2, 5) \), \( A' \) is at \( (4, 8) \)? No, wait, the graph: \( A \) is at \( (2, 5) \), \( A' \) is at \( (4, 8) \)? Wait, no, maybe \( A \) is at \( (2, 5) \), \( A' \) is at \( (4, 8) \): x-shift is \( 4 - 2 = 2 \), y-shift is \( 8 - 5 = 3 \). Wait, but option A is shifted 2 right and 3 up. Wait, but let's check \( B \): \( B \) is at \( (3, 10) \), \( B' \) is at \( (5, 13) \): x-shift \( 5 - 3 = 2 \), y-shift \( 13 - 10 = 3 \). \( C \) is at \( (6, 5) \), \( C' \) is at \( (8, 8) \): x-shift \( 8 - 6 = 2 \), y-shift \( 8 - 5 = 3 \). Wait, but the options are A: 2 right and 3 up, B: 4 right and 3 up. Wait, maybe I misread the coordinates. Wait, maybe \( A \) is at \( (2, 5) \), \( A' \) is at \( (4, 8) \): x difference is 2, y difference is 3. So the translation is 2 units right (x increases by 2) and 3 units up (y increases by 3). So option A.

Wait, but let's recheck. Let's take \( A \): suppose \( A \) is at \( (2, 5) \), \( A' \) is at \( (4, 8) \). So \( x \)-coordinate: \( 4 - 2 = 2 \) (right 2), \( y \)-coordinate: \( 8 - 5 = 3 \) (up 3). So the translation is 2 units right and 3 units up, which is option A.

Step2: Verify with other points

Check \( B \): \( B \) is at \( (3, 10) \), \( B' \) is at \( (5, 13) \). \( 5 - 3 = 2 \) (right 2), \( 13 - 10 = 3 \) (up 3). Check \( C \): \( C \) is at \( (6, 5) \), \( C' \) is at \( (8, 8) \). \( 8 - 6 = 2 \) (right 2), \( 8 - 5 = 3 \) (up 3). So all points have a shift of 2 right and 3 up.

Answer:

A. Shifted 2 units right and 3 units up