Note that you can choose any values for n, t, s, and m.
B
b
Use the same coordinates for the preimage as in Part B. Where does the point end up?
C
c
Where does the point end up after both translations? What coordinates does it have? Is there any difference?
A
a
(x,y)→ (x+n+s,y+t+m)
B
b
(x,y) → (x+s+n,y+m+t)
C
c
No, see solution.
Practice makes perfect
a
Before we do any translations, let's assign values to n, t, s, and m.
n= 2, t= 3, s= 8, m= - 4
Note that these values are completely arbitrary and we could have chosen any values. Using the above values for n and t, we can write translation A.
(x,y) → (x+ n,y+ t)
⇓
(x,y) → (x+ 2,y+ 3)
To make it super easy for us, let's say the preimage is at (0,0). Let's show this translation.
To perform the second translation on the new point, labeled (x',y'), we substitute our chosen value for s and m to write an expression for translation B.
(x',y') → (x'+ s,y'+ t)
⇓
(x',y') → (x'+ 8,y' -4)
Let's show the second translation in our coordinate plane.
To translate the original point to the final image (x'',y'') we have to perform both translations, A and B.
(x,y) → ((x+ 2)+ 8, (y+ 3) -4)
⇕
(x,y) → (x+ 10,y-1)
Let's see now the translation that maps (x,y) onto (x'',y'') on the coordinate plane.
Based on this example, we can write a rule for the combined translation.
(x,y) → (x+ n+ s,y+ t+ m)
b
Let's use the same values as we did in Part A. Let's start with translation B this time.
We will continue with translation A.
Compared to performing translation A first and B second, switching the order does not affect where the final image (x'',y'') ends up.
(x,y) → ((x+ 8)+ 2, (y -4)+ 2)
⇕
(x,y) → (x+ 10,y-1)
Now that we can see how the coordinates of the original point changed based on the two translations, we can reinsert the variables to form an algebraic rule.
(x,y) → (x+ s+ n,y+ m+ t)
c
Now that we have two rules, we need to compare them to see if the order of the translations changes anything.
Part A:& (x,y) → (x+ n+ s,y+ t+ m)
Part B:& (x,y) → (x+ s+ n,y+ m+ t)
Are these the same? Let's check!
