Answer:
A. 8
Explanation:
Point A is located at (-4,4).
If A is translated 6 units to the right and 5 units down, its image A' will be:
[tex]A(-4,4)\to(-4+6,4-5)=A^{\prime}(2,-1)[/tex]Next, find the distance between points A and A':
[tex]\begin{gathered} Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ =\sqrt[]{(2-(-4))^2+(-1-4)^2} \\ =\sqrt[]{(2+4)^2+(-5)^2} \\ =\sqrt[]{(6)^2+(-5)^2} \\ =\sqrt[]{36+25} \\ =\sqrt[]{61} \\ =7.81 \\ \approx8\text{ units} \end{gathered}[/tex]The distance between points A and A' is approximately 8 units.
The correct choice is A.
