Dilation observations worksheet with four problems demonstrating scale factors and coordinate changes.

Let’s solve each problem step by step. We’re dealing with dilations — that means stretching or shrinking a shape from a center point (usually the origin, unless told otherwise) using a scale factor.

We’ll go one problem at a time.

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Problem 1:

Triangle ABC has points:
- A(2, 3)
- B(4, 3)
- C(2, 5)

It is dilated by scale factor ½, centered at the origin.

#### Step 1: Find image coordinates after dilation

To dilate a point (x, y) by scale factor k from the origin, multiply both x and y by k.

So for scale factor ½:

- A'(2 × ½, 3 × ½) = (1, 1.5)
- B'(4 × ½, 3 × ½) = (2, 1.5)
- C'(2 × ½, 5 × ½) = (1, 2.5)

✔ Image coordinates:
A’(1, 1.5), B’(2, 1.5), C’(1, 2.5)

#### Step 2: Draw line through origin and point A

Original point A is (2, 3). The line from origin (0,0) to (2,3) goes up 3, right 2 → slope = 3/2.

After dilation, A’ is (1, 1.5). That’s exactly halfway along the same line! Because 1 is half of 2, and 1.5 is half of 3.

Same for B and C — their images lie on the lines from origin to original points.

#### Step 3: Compare side lengths

Original triangle ABC:

- AB: from (2,3) to (4,3) → horizontal distance = 2 units
- AC: from (2,3) to (2,5) → vertical distance = 2 units
- BC: from (4,3) to (2,5) → use distance formula: √[(4-2)² + (3-5)²] = √[4 + 4] = √8 ≈ 2.83

Image triangle A’B’C’:

- A’B’: from (1,1.5) to (2,1.5) → 1 unit
- A’C’: from (1,1.5) to (1,2.5) → 1 unit
- B’C’: from (2,1.5) to (1,2.5) → √[(2-1)² + (1.5-2.5)²] = √[1 + 1] = √2 ≈ 1.41

Compare:

- AB = 2 → A’B’ = 1 → ratio = 1/2
- AC = 2 → A’C’ = 1 → ratio = 1/2
- BC = √8 → B’C’ = √2 → 2 / √8 = √(2/8) = √(1/4) = 1/2

✔ All sides are scaled by ½.

#### Step 4: What do you notice?

The image triangle is smaller, same shape, and all its vertices lie on the lines connecting the origin to the original vertices. It’s a perfect shrink by half.

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Problem 2:

Segment DE has endpoints:
- D(-2, -1)
- E(2, 1)

Dilated by scale factor 2, centered at origin.

#### Step 1: Find image coordinates

Multiply each coordinate by 2:

- D’(-2×2, -1×2) = (-4, -2)
- E’(2×2, 1×2) = (4, 2)

✔ Image: D’(-4, -2), E’(4, 2)

#### Step 2: Lengths

Original DE: from (-2,-1) to (2,1)

Distance = √[(2 - (-2))² + (1 - (-1))²] = √[4² + 2²] = √[16+4] = √20

Image D’E’: from (-4,-2) to (4,2)

Distance = √[(4 - (-4))² + (2 - (-2))²] = √[8² + 4²] = √[64+16] = √80

Ratio: √80 / √20 = √(80/20) = √4 = 2 ✔

#### Step 3: Notice?

The segment got longer — doubled in length. Also, D’ and E’ lie on the same lines from origin as D and E. For example, D(-2,-1) and D’(-4,-2) are on the same ray from origin — just twice as far.

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Problem 3:

Segment FG has endpoints:
- F(3, 6)
- G(6, 3)

Dilated by scale factor ⅓, centered at origin.

#### Step 1: Image coordinates

Multiply by ⅓:

- F’(3×⅓, 6×⅓) = (1, 2)
- G’(6×⅓, 3×⅓) = (2, 1)

✔ Image: F’(1,2), G’(2,1)

#### Step 2: Lengths

Original FG: from (3,6) to (6,3)

Distance = √[(6-3)² + (3-6)²] = √[9 + 9] = √18

Image F’G’: from (1,2) to (2,1)

Distance = √[(2-1)² + (1-2)²] = √[1 + 1] = √2

Ratio: √2 / √18 = √(2/18) = √(1/9) = 1/3 ✔

#### Step 3: Why doesn’t it look like it changed much?

Because the original segment was already small, and we shrunk it by ⅓ — so it looks almost the same size on the grid. But mathematically, it’s exactly 1/3 the length.

Also, check if F’ lies on line from origin to F:

F is (3,6) → direction vector (3,6) → simplified (1,2) → which is exactly F’! So yes, it’s on the same line, just closer to origin.

Same for G: (6,3) → direction (2,1) → which is G’. Perfect.

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Problem 4:

Triangle XYZ has vertices:
- X(-2, -1)
- Y(-1, 3)
- Z(3, 1)

Dilated by scale factor 2, centered at origin.

#### Step 1: Image coordinates

Multiply each by 2:

- X’(-4, -2)
- Y’(-2, 6)
- Z’(6, 2)

✔ Image: X’(-4,-2), Y’(-2,6), Z’(6,2)

#### Step 2: Perimeter comparison

First, find perimeter of original triangle XYZ.

Use distance formula for each side:

XY: from (-2,-1) to (-1,3)

= √[(-1 - (-2))² + (3 - (-1))²] = √[1² + 4²] = √[1+16] = √17

YZ: from (-1,3) to (3,1)

= √[(3 - (-1))² + (1 - 3)²] = √[4² + (-2)²] = √[16+4] = √20

ZX: from (3,1) to (-2,-1)

= √[(-2 - 3)² + (-1 - 1)²] = √[(-5)² + (-2)²] = √[25+4] = √29

Perimeter original = √17 + √20 + √29

Now image triangle X’Y’Z’:

X’Y’: from (-4,-2) to (-2,6)

= √[(-2 - (-4))² + (6 - (-2))²] = √[2² + 8²] = √[4+64] = √68

Y’Z’: from (-2,6) to (6,2)

= √[(6 - (-2))² + (2 - 6)²] = √[8² + (-4)²] = √[64+16] = √80

Z’X’: from (6,2) to (-4,-2)

= √[(-4 - 6)² + (-2 - 2)²] = √[(-10)² + (-4)²] = √[100+16] = √116

Perimeter image = √68 + √80 + √116

Now compare ratios:

√68 / √17 = √(68/17) = √4 = 2
√80 / √20 = √(80/20) = √4 = 2
√116 / √29 = √(116/29) = √4 = 2

✔ Each side is doubled → perimeter is also doubled.

#### Step 3: Conclusion

When you dilate a figure by scale factor k, every side length becomes k times bigger, so the perimeter also becomes k times bigger.

Here, k=2 → perimeter doubles.

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## Final Answer:

Problem 1:
a. A’(1, 1.5), B’(2, 1.5), C’(1, 2.5)
b. Lines from origin pass through original and image points.
c. Side lengths of image are half the original.
d. Triangle is smaller, same shape, vertices aligned with origin.

Problem 2:
a. D’(-4, -2), E’(4, 2)
b. Original length √20, image length √80 → ratio 2
c. Segment doubled in length, still on same line from origin.

Problem 3:
a. F’(1, 2), G’(2, 1)
b. Original length √18, image length √2 → ratio 1/3
c. Looks similar because shrink is small; but mathematically correct. Points lie on same rays from origin.

Problem 4:
a. X’(-4, -2), Y’(-2, 6), Z’(6, 2)
b. Each side of image is 2× original → perimeter also 2× original.
c. Dilating by 2 doubles all distances from origin → doubles all side lengths → doubles perimeter.