Question

the coordinates of the vertices of △abc are shown in the table below, as well as for △abc and △abc.
△abc △abc △abc
a(1, 3) a(2, 6) a(-6, 6)
b(4, 1) b(8, 2) b(0, 2)
c(5, 6) c(10, 12) c(2, 12)
use the drop - down menus to show what transformations were done to change △abc to △abc.
first △abc was dropdown then △abc was dropdown.

Explanation:

Step1: Analyze transformation to \( \triangle A'B'C' \)

Compare coordinates of \( \triangle ABC \) and \( \triangle A'B'C' \):

• For \( A(1,3) \) to \( A'(2,6) \): \( x \)-coordinate: \( 1\times2 = 2 \), \( y \)-coordinate: \( 3\times2 = 6 \)
• For \( B(4,1) \) to \( B'(8,2) \): \( x \)-coordinate: \( 4\times2 = 8 \), \( y \)-coordinate: \( 1\times2 = 2 \)
• For \( C(5,6) \) to \( C'(10,12) \): \( x \)-coordinate: \( 5\times2 = 10 \), \( y \)-coordinate: \( 6\times2 = 12 \)

So, \( \triangle ABC \) is dilated by a scale factor of 2 (since both \( x \) and \( y \) coordinates are multiplied by 2) to get \( \triangle A'B'C' \).

Step2: Analyze transformation to \( \triangle A''B''C'' \)

Compare coordinates of \( \triangle A'B'C' \) and \( \triangle A''B''C'' \):

• For \( A'(2,6) \) to \( A''(-6,6) \): \( x \)-coordinate: \( 2 - 8 = -6 \), \( y \)-coordinate: \( 6 \) (no change)
• For \( B'(8,2) \) to \( B''(0,2) \): \( x \)-coordinate: \( 8 - 8 = 0 \), \( y \)-coordinate: \( 2 \) (no change)
• For \( C'(10,12) \) to \( C''(2,12) \): \( x \)-coordinate: \( 10 - 8 = 2 \), \( y \)-coordinate: \( 12 \) (no change)

So, \( \triangle A'B'C' \) is translated 8 units to the left (since \( x \)-coordinate decreases by 8, \( y \)-coordinate remains same) to get \( \triangle A''B''C'' \).

Answer:

First \( \triangle ABC \) was \(\boldsymbol{\text{dilated}}\) (by scale factor 2), then \( \triangle A'B'C' \) was \(\boldsymbol{\text{translated}}\) (8 units left).