We will begin by drawing the two triangles in a coordinate plane.
We can immediately see that the figures have different orientations. Therefore, to map △ ABC to △ DEF, we have to rotate △ ABC a certain number of degrees about the origin. To determine this rotation, we will compare the slopes of two corresponding sides such as DF and AC.
| Side
|
Points
|
y_2-y_1/x_2-x_1
|
m
|
| DF
|
( 2,- 1), ( - 1,2)
|
- 1- 2/2-( - 1)
|
- 1
|
| AC
|
( 4,4), ( 1,1)
|
4- 1/4- 1
|
1
|
The slope of DF and of AC are negative reciprocals of each other which means these sides are perpendicular. To align the orientation of perpendicular sides, we can for example rotate △ ABC by 90^(∘).
When we rotate a figure by 90^(∘) counterclockwise about the origin, the coordinates of the vertices change in the following way
preimage (a,b) → image (- b,a)
Let's perform this transformation on the triangles vertices.
| Point
|
(a,b)
|
(- b,a)
|
| A
|
(1,1)
|
(- 1,1)
|
| B
|
(3,2)
|
(- 2,3)
|
| C
|
(4,4)
|
(- 4,4)
|
Now we can draw △ A'B'C'.
Even though A'C' now have the same orientation as DF, the remaining two sides of the triangles do not have the same orientation. Instead, we will rotate △ ABC by 270^(∘) which will align all of the sides.
When a figure is rotated by 270^(∘) counterclockwise about the origin, the coordinates of the vertices change in the following way
preimage (a,b) → image (b,- a)
Let's perform this transformation on the vertices of △ ABC.
| Point
|
(a,b)
|
(b,- a)
|
| A
|
(1,1)
|
(1,- 1)
|
| B
|
(3,2)
|
(2,- 3)
|
| C
|
(4,4)
|
(4,- 4)
|
Now we can draw △ A'B'C'.
When the triangles have the same orientation, we finally have to perform a translation to map △ A'B'C' to △ DEF. The vertices C' and D are corresponding vertices. Therefore, we have to translate △ A'B'C' by 3 units up and 2 units to the left for them to map to each other.
A composition of rigid motions that maps △ ABC to △ DEF is:
Rotation:& 270^(∘) about the origin.
Translation:& (x,y) → (x-2,y+3).
