Translations on the Coordinate Plane Worksheet | PDF Printable ... - Free Printable

Let's solve the "Translation Trail" worksheet step by step. This is a geometry exercise involving translations on the coordinate plane.

We are given a grid with labeled points (A, B, C, ..., Z) and asked to:

1. Find the translation vector (as a column vector) or describe it in words for given point movements.
2. Fill in the table with either:
- Words → Vector
- Vector → Words

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🔍 Step 1: Understand Coordinates

Each point has coordinates (x, y). We’ll read them from the grid.

Let’s first locate key points:

| Point | Coordinates |
|-------|-------------|
| A | (4, 2) |
| O | (2, 5) |
| R | (-4, 8) |
| S | (1, -4) |
| C | (-6, 1) |
| I | (-5, -7) |
| H | (-8, 5) |
| T | (-7, -4) |
| U | (4, -5) |

Now let’s fill in the table.

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## ✔ Section A: Complete the Table

Part 1: From Point to Point (A to O, etc.)

#### 1. A to O
- A = (4, 2)
- O = (2, 5)
- Change in x: 2 – 4 = -2
- Change in y: 5 – 2 = +3
- Vector: $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$
- Words: 2 left, 3 up

✔ Answer:
Words: 2 left, 3 up
Vector: $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$

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#### 2. R to S
- R = (-4, 8)
- S = (1, -4)
- Δx = 1 – (-4) = +5
- Δy = -4 – 8 = -12
- Vector: $\begin{pmatrix} 5 \\ -12 \end{pmatrix}$
- Words: 5 right, 12 down

✔ Answer:
Words: 5 right, 12 down
Vector: $\begin{pmatrix} 5 \\ -12 \end{pmatrix}$

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#### 3. C to I
- C = (-6, 1)
- I = (-5, -7)
- Δx = -5 – (-6) = +1
- Δy = -7 – 1 = -8
- Vector: $\begin{pmatrix} 1 \\ -8 \end{pmatrix}$
- Words: 1 right, 8 down

✔ Answer:
Words: 1 right, 8 down
Vector: $\begin{pmatrix} 1 \\ -8 \end{pmatrix}$

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#### 4. H to T
- H = (-8, 5)
- T = (-7, -4)
- Δx = -7 – (-8) = +1
- Δy = -4 – 5 = -9
- Vector: $\begin{pmatrix} 1 \\ -9 \end{pmatrix}$
- Words: 1 right, 9 down

✔ Answer:
Words: 1 right, 9 down
Vector: $\begin{pmatrix} 1 \\ -9 \end{pmatrix}$

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#### 5. A to U
- A = (4, 2)
- U = (4, -5)
- Δx = 4 – 4 = 0
- Δy = -5 – 2 = -7
- Vector: $\begin{pmatrix} 0 \\ -7 \end{pmatrix}$
- Words: 7 down

✔ Answer:
Words: 7 down
Vector: $\begin{pmatrix} 0 \\ -7 \end{pmatrix}$

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Part 2: From Words or Vectors to Points

Now fill in the right-hand side of the table.

#### 1. 3 right
- This means move +3 in x-direction.
- So any point moved 3 right will have vector: $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$
- Example: Let's find two points where this applies.

But we need to find which translation corresponds — likely asking for a pair of points that satisfy this.

Let’s look at the grid.

Try: Q to D?
Q = (0, -8), D = (7, -7)? No.

Wait — better idea: Look for a pair where x increases by 3.

Try P to ?
P = (-3, -2) → if go 3 right → x = 0, y = -2 → point at (0, -2)? Not labeled.

Try E to ?
E = (-2, 1) → 3 right → (1, 1) → not labeled.

Wait — how about Z to ?
Z = (-4, 0) → 3 right → (-1, 0) → no label.

Try N to ?
N = (-9, -3) → 3 right → (-6, -3) → not labeled.

Hmm. Maybe try M to ?
M = (-9, -9) → 3 right → (-6, -9) → not labeled.

Alternatively, maybe I to ?
I = (-5, -7) → 3 right → (-2, -7) → no.

Wait — what about K to ?
K = (8, 0) → 3 right → (11, 0) → off grid.

Wait — perhaps L to ?
L = (3, 6) → 3 right → (6, 6) → not labeled.

Wait — maybe J to ?
J = (2, -9) → 3 right → (5, -9) → not labeled.

No obvious one.

But the question says: “Complete the table.” It may be asking for any valid pair that satisfies the translation.

So instead of guessing, let’s think: The vector for "3 right" is $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$

So we can write:

- Words: 3 right
- Vector: $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$

And now pick two points such that moving from one to the other is 3 right.

For example:

- From G to W: G = (-9, 9), W = (9, 9) → too far.

Try F to ?
F = (7, 4) → 3 right → (10, 4) → not labeled.

Wait — what about B to ?
B = (-3, 5) → 3 right → (0, 5) → no label.

How about Y to L? Y = (1, 7), L = (3, 6) → not 3 right.

Wait — maybe S to ?
S = (1, -4) → 3 right → (4, -4) → not labeled.

Wait — maybe U to ?
U = (4, -5) → 3 right → (7, -5) → not labeled.

Hmm.

Wait — what about D to X?
D = (7, -7), X = (6, -9) → no.

Wait — maybe X to ?
X = (6, -9) → 3 right → (9, -9) → not labeled.

Wait — perhaps O to L?
O = (2, 5), L = (3, 6) → not 3 right.

Wait — maybe P to ?
P = (-3, -2) → 3 right → (0, -2) → no.

Wait — perhaps T to ?
T = (-7, -4) → 3 right → (-4, -4) → no.

Wait — what about C to ?
C = (-6, 1) → 3 right → (-3, 1) → not labeled.

Wait — E to ?
E = (-2, 1) → 3 right → (1, 1) → no.

Wait — maybe Z to ?
Z = (-4, 0) → 3 right → (-1, 0) → no.

None seem to match.

Wait — what about Q to ?
Q = (0, -8) → 3 right → (3, -8) → not labeled.

Wait — V = (3, -8) → so Q = (0, -8), V = (3, -8) → yes!

So Q to V is 3 right.

Q = (0, -8), V = (3, -8) → Δx = +3, Δy = 0 → perfect.

✔ So Q to V → 3 right → vector $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$

Answer:
From: Q to V
Words: 3 right
Vector: $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$

So we can write:
Q to V

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#### 2. 1 left, 4 up

- Left: x decreases by 1 → -1
- Up: y increases by 4 → +4
- Vector: $\begin{pmatrix} -1 \\ 4 \end{pmatrix}$

Find a pair of points that satisfy this.

Try:
Look for point A = (4, 2) → move 1 left, 4 up → (3, 6) → that's L

L = (3, 6)

So A to L?

A = (4, 2), L = (3, 6): Δx = -1, Δy = +4 → yes!

✔ So A to L

Answer:
From: A to L
Words: 1 left, 4 up
Vector: $\begin{pmatrix} -1 \\ 4 \end{pmatrix}$

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#### 3. Vector $\begin{pmatrix} 11 \\ 0 \end{pmatrix}$

This means 11 units right, 0 up/down.

So we need a point that moves 11 units right to another point.

Look for two points with same y-coordinate, and x differs by 11.

Check possible points:

- M = (-9, -9), X = (6, -9) → difference = 15 → too much
- N = (-9, -3), D = (7, -7) → different y
- G = (-9, 9), W = (9, 9) → x diff = 18 → too much
- C = (-6, 1), F = (7, 4) → not same y

Wait — G = (-9, 9), W = (9, 9) → x diff = 18

No.

What about Z = (-4, 0), K = (8, 0) → diff = 12 → close

Wait — H = (-8, 5), F = (7, 4) → not same y

Wait — M = (-9, -9), J = (2, -9) → x diff = 11 → yes!

M = (-9, -9), J = (2, -9) → Δx = 2 – (-9) = 11, Δy = 0 → perfect!

So M to J → $\begin{pmatrix} 11 \\ 0 \end{pmatrix}$

✔ Answer: M to J

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#### 4. Vector $\begin{pmatrix} -3 \\ 5 \end{pmatrix}$

Means: 3 left, 5 up

Find a point that moves 3 left and 5 up to another.

Try:
Start from a point with x ≥ 3, y ≤ something.

Try B = (-3, 5) → move 3 left → x = -6, 5 up → y = 10 → (-6, 10) → no label

Try P = (-3, -2) → 3 left → (-6, -2), 5 up → (-6, 3) → no

Try T = (-7, -4) → 3 left → (-10, -4), 5 up → (-10, 1) → no

Try R = (-4, 8) → 3 left → (-7, 8), 5 up → (-7, 13) → off grid

Try Y = (1, 7) → 3 left → (-2, 7), 5 up → (-2, 12) → no

Wait — D = (7, -7) → 3 left → (4, -7), 5 up → (4, -2) → not labeled

Wait — O = (2, 5) → 3 left → (-1, 5), 5 up → (-1, 10) → no

Wait — S = (1, -4) → 3 left → (-2, -4), 5 up → (-2, 1) → is there a point at (-2, 1)? Yes! That’s E

E = (-2, 1)

S = (1, -4), E = (-2, 1)

Δx = -2 – 1 = -3 → 3 left
Δy = 1 – (-4) = +5 → 5 up

Perfect!

So S to E → $\begin{pmatrix} -3 \\ 5 \end{pmatrix}$

✔ Answer: S to E

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#### 5. Vector $\begin{pmatrix} 2 \\ -7 \end{pmatrix}$

Means: 2 right, 7 down

Find such a move.

Try:
Start from a point with x ≤ 8, y ≥ 7

Try W = (9, 9) → 2 right → (11, 9) → off grid

Try F = (7, 4) → 2 right → (9, 4), 7 down → (9, -3) → not labeled

Try A = (4, 2) → 2 right → (6, 2), 7 down → (6, -5) → not labeled

Try L = (3, 6) → 2 right → (5, 6), 7 down → (5, -1) → no

Try Y = (1, 7) → 2 right → (3, 7), 7 down → (3, 0) → no

Try H = (-8, 5) → 2 right → (-6, 5), 7 down → (-6, -2) → not labeled

Try C = (-6, 1) → 2 right → (-4, 1), 7 down → (-4, -6) → no

Wait — G = (-9, 9) → 2 right → (-7, 9), 7 down → (-7, 2) → not labeled

Wait — B = (-3, 5) → 2 right → (-1, 5), 7 down → (-1, -2) → not labeled

Wait — P = (-3, -2) → 2 right → (-1, -2), 7 down → (-1, -9) → no

Wait — N = (-9, -3) → 2 right → (-7, -3), 7 down → (-7, -10) → off grid

Wait — T = (-7, -4) → 2 right → (-5, -4), 7 down → (-5, -11) → off

Wait — Q = (0, -8) → 2 right → (2, -8), 7 down → (2, -15) → no

Wait — J = (2, -9) → 2 right → (4, -9), 7 down → (4, -16) → no

Wait — U = (4, -5) → 2 right → (6, -5), 7 down → (6, -12) → no

Wait — D = (7, -7) → 2 right → (9, -7), 7 down → (9, -14) → no

Wait — K = (8, 0) → 2 right → (10, 0), 7 down → (10, -7) → off

Wait — maybe X = (6, -9) → 2 right → (8, -9), 7 down → (8, -16) → no

Wait — I = (-5, -7) → 2 right → (-3, -7), 7 down → (-3, -14) → no

Wait — P = (-3, -2) → 2 right → (-1, -2), 7 down → (-1, -9) → not labeled

Wait — Z = (-4, 0) → 2 right → (-2, 0), 7 down → (-2, -7) → is there a point at (-2, -7)? Yes! That’s I

I = (-5, -7)? Wait no — earlier we said I = (-5, -7)

Wait — check grid: I is at (-5, -7)? Let's recheck.

Looking back:
On grid: I is at x = -5, y = -7 → yes.

But (-2, -7) is not labeled.

Wait — what about P = (-3, -2) → 2 right → (-1, -2), 7 down → (-1, -9) → no

Wait — Q = (0, -8) → 2 right → (2, -8), 7 down → (2, -15) → no

Wait — J = (2, -9) → 2 right → (4, -9), 7 down → (4, -16) → no

Wait — V = (3, -8) → 2 right → (5, -8), 7 down → (5, -15) → no

Wait — O = (2, 5) → 2 right → (4, 5), 7 down → (4, -2) → is there a point at (4, -2)? No.

Wait — A = (4, 2) → 2 right → (6, 2), 7 down → (6, -5) → no

Wait — F = (7, 4) → 2 right → (9, 4), 7 down → (9, -3) → no

Wait — D = (7, -7) → 2 right → (9, -7), 7 down → (9, -14) → no

Wait — W = (9, 9) → 2 right → (11, 9) → off

Wait — maybe R = (-4, 8) → 2 right → (-2, 8), 7 down → (-2, 1) → that’s E

E = (-2, 1)

R = (-4, 8) → 2 right → (-2, 8), then 7 down → (-2, 1) → yes!

So R to E?

R = (-4, 8), E = (-2, 1)

Δx = -2 – (-4) = +2
Δy = 1 – 8 = -7 → yes!

So R to E → $\begin{pmatrix} 2 \\ -7 \end{pmatrix}$

✔ Answer: R to E

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## ✔ Final Answers

Left Table (Point to Point):

| From | Words | Vector |
|------|--------|--------|
| A to O | 2 left, 3 up | $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ |
| R to S | 5 right, 12 down | $\begin{pmatrix} 5 \\ -12 \end{pmatrix}$ |
| C to I | 1 right, 8 down | $\begin{pmatrix} 1 \\ -8 \end{pmatrix}$ |
| H to T | 1 right, 9 down | $\begin{pmatrix} 1 \\ -9 \end{pmatrix}$ |
| A to U | 7 down | $\begin{pmatrix} 0 \\ -7 \end{pmatrix}$ |

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Right Table (Words or Vector to Points):

| From | Words | Vector |
|------|--------|--------|
| Q to V | 3 right | $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$ |
| A to L | 1 left, 4 up | $\begin{pmatrix} -1 \\ 4 \end{pmatrix}$ |
| M to J | 11 right | $\begin{pmatrix} 11 \\ 0 \end{pmatrix}$ |
| S to E | 3 left, 5 up | $\begin{pmatrix} -3 \\ 5 \end{pmatrix}$ |
| R to E | 2 right, 7 down | $\begin{pmatrix} 2 \\ -7 \end{pmatrix}$ |

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✔ All filled in correctly.

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