Composition of Transformations Practice - MathBitsNotebook(Geo) - Free Printable

Looking at the image, we can see two triangles on a coordinate plane:

- Triangle BUG in the second quadrant (top left)
- Triangle B’U’G’ in the fourth quadrant (bottom right)

We are likely being asked to describe the transformation that maps triangle BUG onto triangle B’U’G’.

Let’s find the coordinates of each point:

Triangle BUG:
- B is at (-6, 1)
- U is at (-3, 7)
- G is at (-2, 1)

Triangle B’U’G’:
- B’ is at (0, -6)
- U’ is at (6, -3)
- G’ is at (0, -2)

Wait — let me double-check those coordinates from the grid.

Actually, looking again:

For triangle BUG:
- Point B: x = -6, y = 1 → (-6, 1)
- Point U: x = -3, y = 7 → (-3, 7)
- Point G: x = -2, y = 1 → (-2, 1)

For triangle B’U’G’:
- Point B’: x = 0, y = -6 → (0, -6)
- Point U’: x = 6, y = -3 → (6, -3)
- Point G’: x = 0, y = -2 → (0, -2)

Now, let’s see how each point moves:

From B(-6, 1) to B’(0, -6):
- x: -6 → 0 → change of +6
- y: 1 → -6 → change of -7

That doesn’t look like a simple translation. Let’s check if it’s a rotation.

Try rotating 90° clockwise around origin:
Rule: (x, y) → (y, -x)

B(-6, 1) → (1, 6) → not matching B’(0, -6)

Try 90° counterclockwise: (x, y) → (-y, x)
B(-6, 1) → (-1, -6) → not matching

Try 180°: (x, y) → (-x, -y)
B(-6, 1) → (6, -1) → not matching B’(0, -6)

Hmm. Maybe it’s a reflection?

What if we reflect over the line y = -x?
Rule: (x, y) → (-y, -x)

B(-6, 1) → (-1, 6) → no

Wait — let’s try reflecting over the origin? That’s same as 180° rotation.

Alternatively, maybe it’s a combination? Or perhaps I misread the points.

Let me recheck the image carefully.

Actually, looking again at the image:

Triangle BUG:
- B is at (-6, 1) — correct
- U is at (-3, 7) — correct
- G is at (-2, 1) — correct

Triangle B’U’G’:
- B’ is at (0, -6) — yes
- U’ is at (6, -3) — yes
- G’ is at (0, -2) — yes

Wait — what if we consider mapping B to G’? No, labels should correspond.

Perhaps it’s a rotation about a different point?

Let’s try to see vector from B to B’:
B(-6,1) to B’(0,-6): Δx = +6, Δy = -7

U(-3,7) to U’(6,-3): Δx = +9, Δy = -10 — not same.

Not a translation.

Maybe it’s a rotation of 90 degrees clockwise but around a different center?

Alternatively, let’s try to see if it’s a reflection over the line y = x?

Reflection over y=x: (x,y) → (y,x)

B(-6,1) → (1,-6) — close to B’(0,-6)? Not quite.

Wait — what if it’s a reflection over the line y = -x?

(x,y) → (-y, -x)

B(-6,1) → (-1, 6) — no

Another idea: perhaps it’s a 90-degree clockwise rotation followed by a translation? Too complicated.

Let me plot mentally:

Original triangle BUG has base BG horizontal from x=-6 to x=-2 at y=1, and peak U at (-3,7).

Image triangle B’U’G’ has points at (0,-6), (6,-3), (0,-2). So it looks like it's oriented differently.

Notice that in BUG, B and G have same y-coordinate (y=1), so base is horizontal.

In B’U’G’, B’ and G’ have same x-coordinate (x=0), so side B’G’ is vertical.

That suggests a 90-degree rotation.

Let’s assume rotation of 90 degrees clockwise around origin: (x,y) → (y, -x)

Apply to B(-6,1): (1, 6) — but B’ is (0,-6) — not match.

What if we rotate 90 degrees counterclockwise: (x,y) → (-y, x)

B(-6,1) → (-1, -6) — still not (0,-6)

Close! (-1,-6) vs (0,-6) — off by 1 in x.

Try U(-3,7) → (-7, -3) — but U’ is (6,-3) — not close.

This isn't working.

Alternative approach: maybe it's a reflection over the line y = -x + c or something.

Let’s calculate midpoints between corresponding points to find center of rotation or reflection.

If it's a rotation, the center is equidistant from preimage and image.

Assume B maps to B’, U to U’, G to G’.

Midpoint of B and B’: ((-6+0)/2, (1+(-6))/2) = (-3, -2.5)

Midpoint of U and U’: ((-3+6)/2, (7+(-3))/2) = (1.5, 2)

Different midpoints — so not a point reflection (180° rotation).

For 90° rotation, the center would be such that vectors are perpendicular.

Let me try to find a point C(h,k) such that rotating B around C gives B’.

Rotation 90° clockwise:
x' = h + (y - k)
y' = k - (x - h)

So for B(-6,1) to become B’(0,-6):

0 = h + (1 - k) => h - k = -1 ...(1)
-6 = k - (-6 - h) => -6 = k +6 + h => h + k = -12 ...(2)

Add (1) and (2): 2h = -13 => h = -6.5
Then from (1): -6.5 - k = -1 => k = -5.5

Now check with U(-3,7) rotating around (-6.5, -5.5) 90° clockwise:

x' = h + (y - k) = -6.5 + (7 - (-5.5)) = -6.5 + 12.5 = 6
y' = k - (x - h) = -5.5 - (-3 - (-6.5)) = -5.5 - (3.5) = -9

But U’ is (6, -3), not (6,-9). So not matching.

Try 90° counterclockwise around C(h,k):

x' = h - (y - k)
y' = k + (x - h)

For B(-6,1) to B’(0,-6):

0 = h - (1 - k) => h -1 + k = 0 => h + k = 1 ...(1)
-6 = k + (-6 - h) => -6 = k -6 - h => -h + k = 0 => k = h ...(2)

From (2) k=h, plug into (1): h + h =1 => 2h=1 => h=0.5, k=0.5

Check U(-3,7) around (0.5,0.5) 90° CCW:

x' = 0.5 - (7 - 0.5) = 0.5 - 6.5 = -6
y' = 0.5 + (-3 - 0.5) = 0.5 - 3.5 = -3

So U' would be (-6, -3), but actual U’ is (6,-3). Close but x is wrong sign.

Not matching.

Perhaps it's a reflection.

Let me try reflecting over the line y = -x.

Rule: (x,y) -> (-y, -x)

B(-6,1) -> (-1,6) — not B’(0,-6)

Reflect over y = x: (x,y)->(y,x)

B(-6,1)->(1,-6) — close to B’(0,-6)? Off by 1.

U(-3,7)->(7,-3) — but U’ is (6,-3) — off by 1 in x.

G(-2,1)->(1,-2) — but G’ is (0,-2) — off by 1 in x.

Oh! What if we reflect over y=x and then shift left by 1? Or something.

Notice that after reflection over y=x, we get:
B(1,-6), U(7,-3), G(1,-2)

But actual image is B’(0,-6), U’(6,-3), G’(0,-2)

So each x-coordinate is decreased by 1.

So transformation: reflect over y=x, then translate left by 1 unit.

Reflect over y=x: (x,y) -> (y,x)
Then translate left by 1: (x,y) -> (x-1, y)

So overall: (x,y) -> (y-1, x)

Check for B(-6,1): (1-1, -6) = (0, -6) — matches B’!

U(-3,7): (7-1, -3) = (6, -3) — matches U’!

G(-2,1): (1-1, -2) = (0, -2) — matches G’!

Perfect!

So the transformation is: reflect over the line y = x, then translate 1 unit to the left.

We can write it as a single rule: (x, y) → (y - 1, x)

But typically, we describe transformations as compositions.

Since the problem likely expects a description, we can say:

The triangle is reflected over the line y = x and then translated 1 unit to the left.

Alternatively, since the net effect is (x,y) -> (y-1, x), but that's not a standard named transformation.

In many curricula, they might expect "reflection over y=x followed by translation left 1".

But let me confirm if there's a simpler way.

Notice that (x,y) -> (y-1, x) can also be seen as a rotation or other, but it's not a pure rotation or reflection.

Another way: it's equivalent to a glide reflection or something, but probably overcomplicating.

Given that, I think the intended answer is reflection over y=x followed by translation left 1 unit.

But let me see the positions again.

Perhaps it's a rotation of 90 degrees counterclockwise around the point (0.5, 0.5) but earlier calculation didn't work.

Earlier when I tried 90° CCW around (0.5,0.5), I got U' at (-6,-3) but should be (6,-3), so not.

Unless I made a mistake.

Perhaps the correspondence is different. Maybe B maps to G' or something.

Let me try assuming B maps to G'.

B(-6,1) to G'(0,-2): Δx=6, Δy=-3

U(-3,7) to U'(6,-3): Δx=9, Δy=-10 — not same.

Or B to U': (-6,1) to (6,-3): Δx=12, Δy=-4

Not consistent.

So probably the labeling is correct: B to B', U to U', G to G'.

And we found that (x,y) -> (y-1, x) works.

We can describe it as: first reflect over the line y = x, then move 1 unit left.

In some contexts, they might call this a "glide reflection" but glide reflection is usually reflection over a line followed by translation parallel to that line. Here, reflection over y=x, then translation left (which is not parallel to y=x), so not a glide reflection.

So best to describe as composition.

Perhaps the problem is to identify the type of transformation, but since it's not a single rigid motion, it's a composition.

But looking back at the image, and the fact that the ant is there, perhaps it's just to recognize the mapping.

Another thought: maybe it's a 90-degree rotation around the origin but I miscalculated the points.

Let me list coordinates again from the grid:

Grid lines: each square is 1 unit.

Triangle BUG:
- B: x=-6, y=1 → (-6,1)
- U: x=-3, y=7 → (-3,7)
- G: x=-2, y=1 → (-2,1)

Triangle B’U’G’:
- B’: x=0, y=-6 → (0,-6)
- U’: x=6, y=-3 → (6,-3)
- G’: x=0, y=-2 → (0,-2)

Now, let's compute the vector from B to B': (0 - (-6), -6 - 1) = (6, -7)

From U to U': (6 - (-3), -3 - 7) = (9, -10)

From G to G': (0 - (-2), -2 - 1) = (2, -3)

Not the same, so not translation.

Distances:

In BUG:
BG: from (-6,1) to (-2,1) = distance 4
BU: from (-6,1) to (-3,7) = sqrt((3)^2 + (6)^2) = sqrt(9+36)=sqrt(45)=3sqrt(5)
GU: from (-2,1) to (-3,7) = sqrt((-1)^2 + (6)^2)=sqrt(1+36)=sqrt(37)

In B’U’G’:
B’G’: from (0,-6) to (0,-2) = distance 4
B’U’: from (0,-6) to (6,-3) = sqrt(6^2 + 3^2)=sqrt(36+9)=sqrt(45)=3sqrt(5)
G’U’: from (0,-2) to (6,-3) = sqrt(6^2 + (-1)^2)=sqrt(36+1)=sqrt(37)

Same side lengths, so congruent, good.

Now, orientation: in BUG, going from B to U to G, it's clockwise or counterclockwise?

B(-6,1), U(-3,7), G(-2,1)

Vector BU = (3,6), vector BG = (4,0)

Cross product BU x BG = 3*0 - 6*4 = -24 <0, so clockwise.

In B’U’G’: B’(0,-6), U’(6,-3), G’(0,-2)

Vector B’U’ = (6,3), vector B’G’ = (0,4)

Cross product 6*4 - 3*0 = 24 >0, so counterclockwise.

So orientation reversed, which means it's an opposite isometry, like reflection or glide reflection.

Since it's not a single reflection (as we saw), it must be a glide reflection or composition.

But earlier we found a composition that works: reflect over y=x then translate left 1.

Reflecting over y=x reverses orientation, and translation preserves it, so overall reverses orientation, which matches.

And we verified the points.

So the transformation is: reflection over the line y = x followed by a translation of 1 unit to the left.

We can write the rule as (x, y) → (y - 1, x)

But for the answer, since the problem doesn't specify what to provide, but from context, likely they want the description of the transformation.

Perhaps in the original problem, it's to identify the type, but since it's not specified, and based on common tasks, I think describing the transformation is expected.

Another possibility: maybe it's a rotation of 90 degrees clockwise around the point (0.5, -0.5) or something, but let's not overcomplicate.

I think the simplest is to state the composition.

So, final answer should be: The triangle is reflected over the line y = x and then translated 1 unit to the left.

To box it.

But let me see if there's a single transformation.

Notice that (x,y) -> (y-1, x) can be written as a matrix or something, but for school level, composition is fine.

Perhaps they consider it as a rotation, but it's not.

Another idea: what if we rotate 90 degrees counterclockwise around (0.5, 0.5), but earlier calculation gave wrong point.

Let me recalculate rotation 90° CCW around (h,k):

x' = h - (y - k)
y' = k + (x - h)

Set for B(-6,1) to B'(0,-6):

0 = h - (1 - k) => h -1 + k = 0 => h + k = 1

-6 = k + (-6 - h) => -6 = k -6 - h => -h + k = 0 => k = h

Then h + h =1 => h=0.5, k=0.5

Then for U(-3,7): x' = 0.5 - (7 - 0.5) = 0.5 - 6.5 = -6

y' = 0.5 + (-3 - 0.5) = 0.5 - 3.5 = -3

So (-6, -3), but U' is (6,-3), so if we take absolute value or something, no.

Unless the correspondence is different. Suppose U maps to G' or something.

Let me try B to B', U to G', G to U'.

B(-6,1) to B'(0,-6): as before.

U(-3,7) to G'(0,-2): Δx=3, Δy=-9

G(-2,1) to U'(6,-3): Δx=8, Δy=-4 — not consistent.

B to G': (-6,1) to (0,-2): Δx=6, Δy=-3

U to U': (-3,7) to (6,-3): Δx=9, Δy=-10

G to B': (-2,1) to (0,-6): Δx=2, Δy=-7 — no.

So probably the labeling is correct.

Perhaps it's a reflection over the line y = -x + c.

Let me assume reflection over y = -x + b.

The formula for reflection over y = -x + b is a bit messy.

General formula for reflection over ax+by+c=0, but complicated.

Since we already have a working composition, and it fits perfectly, I'll go with that.

So the transformation is reflection over the line y = x followed by a translation of 1 unit to the left.

Final Answer: \boxed{\text{reflection over the line } y = x \text{ followed by a translation of 1 unit to the left}}

To make it neat.

Some might write "reflect over y=x, then shift left 1".

But I think that's fine.

Perhaps they want the net rule, but I think description is better.

Another way: since (x,y) -> (y-1, x), we can say it's equivalent to swapping x and y and then subtracting 1 from the new x-coordinate, but same thing.

I think for the answer, we'll put the description.

So, after careful analysis, the transformation that maps triangle BUG to triangle B'U'G' is a reflection over the line y = x followed by a translation 1 unit to the left.