AA'B'C' is the image of AABC under a translation71Aనా-----6+A5+4+32CU14《一7-6-5 -4 -3 2 1 2 3 4 5 6 72B-3+-4В-5-6+a>

To determine the translation applied to determine ΔA'B'C' starting from ΔABC you have to compare the coordinates of two corresponding vertices, for example, A and A'
The coordinates of vertex A are (3,5)
The coordinates of vertex A' are (7,7)
ΔA'B'C' is to the right of ΔABC, which suggests that the triangle was moved a determined number "k" of units to the right. To perform this horizontal translation, "k" units were added to the x-coordinate of each point.
For points A and A', this means that to determine the x-coordinate of A', k units were added to the x-coordinate of A:
[tex]x_A+k=x_{A^{\prime}}_{}[/tex]Using this expression we can determine the units the triangle was moved to the right:
[tex]3+k=7[/tex]-Subtract 3 from both sides:
[tex]\begin{gathered} 3-3+k=7-3 \\ k=4 \end{gathered}[/tex]k=4 indicates that ΔABC was translated 4 units to the right
ΔA'B'C' is above ΔABC, which suggests that it was translated vertically upwards.
To perform a vertical translation up, you have to add n units to the y-coordinate of each point.
Comparing points A and A', when the triangle was translated vertically, there were "n" units added to the y-coordinate of A so that:
[tex]y_A+n=y_{A^{\prime}}[/tex]Using the y-coordinates of both vertices you can determine the value of "n"
[tex]5+n=7[/tex]-Subtract 5 from both sides:
[tex]\begin{gathered} 5-5+n=7-5 \\ n=2 \end{gathered}[/tex]n=2 indicates that ΔABC was translated 2 units up
You can write the general rule used to translate ΔABC as follows:
[tex](x,y)\to(x+4,y+2)[/tex]