Answer:
Part A: The x-coordinates of points A, B , C are add or subtract by 6
Part B: The y-coordinates of points A, B , C are add or subtract by 3
Part C: A' = (-4 ± 6 , 1 ± 3) , B' = (-1 ± 6 , 1 ± 3) , C' = (-1 ± 6 , 4 ± 3)
Step-by-step explanation:
* Lets explain the answer
- If a point (x , y) is translated horizontally h units to the right, then
its image is (x + h , y)
- If a point (x , y) is translated horizontally h units to the left, then
its image is (x - h , y)
- If a point (x , y) is translated vertically k units up, then its image
is (x , y + k)
- If a point (x , y) is translated vertically k units down, then its image
is (x , y - k)
* Lets solve the problem
- Triangle ABC has vertices A = (-4 , 1) , B = (-1 , 1) , C = (-1 , 4)
# Part A:
∵ Δ ABC translated 6 units horizontally
∴ The x-coordinates of points A, B , C are add or subtract by 6
- If the triangle translated to the right then each x-coordinate of
points A, B , C added by 6
- If the triangle translated to the left then each x-coordinate of
points A, B , C subtracted by 6
# Part B:
∵ Δ ABC translated 3 units vertically
∴ The y-coordinates of points A, B , C are add or subtract by 3
- If the triangle translated up then each y-coordinate of points
A, B , C added by 3
- If the triangle translated down then each y-coordinate of points
A, B , C subtracted by 3
# Part C
∵ The vertices of Δ ABC are A = (-4 , 1) , B = (-1 , 1) , C = (-1 , 4)
- The triangle translated 6 units horizontally and 3 units vertically
∴ The images of its vertices are:
A' = (-4 ± 6 , 1 ± 3) , B' = (-1 ± 6 , 1 ± 3) , C' = (-1 ± 6 , 4 ± 3)
