Dilation: Scaling from a Center Point

A dilation scales every point uniformly outward (or inward) from a fixed center point. Unlike rotations and reflections, dilations change the size of figures while preserving the shape.

The Formula

For a dilation by scale factor $k$ centered at the origin:

$(x, y) \to (kx,\, ky)$

For a center other than the origin, first translate so the center is at the origin, dilate, then translate back.

The scale factor $k$ controls everything:

• $k > 1$ → enlargement (figure grows)
• $0 < k < 1$ → reduction (figure shrinks)
• $k = 1$ → identity (no change)
• $k < 0$ → enlargement and a 180° rotation (figure flips)

How Dimensions Scale

For a dilation by factor $k$:

• Lengths scale by $k$
• Areas scale by $k^{2}$
• Volumes scale by $k^{3}$ (in 3D)

This is the same as for similar figures — and indeed, a dilated figure is similar to the original.

Step-by-Step Solution

Given: A square with vertices $(1, 1)$, $(3, 1)$, $(3, 3)$, $(1, 3)$. Dilate from the origin with scale factor $k = 2$.

Find: The image vertices, the new side length, and the new area.

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Step 1 — Apply $(x, y) \to (2x, 2y)$ to each vertex.

$(1, 1) \to (2, 2)$

$(3, 1) \to (6, 2)$

$(3, 3) \to (6, 6)$

$(1, 3) \to (2, 6)$

Step 2 — Verify by computing side lengths.

Original side length: $3 - 1 = 2$ → side = 2.

Image side length: $6 - 2 = 4$ → side = 4.

Side length doubled, as expected ($k = 2$).

Step 3 — Compute the original area.

$A_1 = 2 \times 2 = 4$

Step 4 — Compute the image area.

$A_2 = 4 \times 4 = 16$

The area is 4 times the original ($k^{2} = 4$). Doubling the linear dimension quadruples the area.

Step 5 — Verify the figure is similar to the original.

Both squares have:

• 4 right angles
• 4 equal sides (4 times the original)
• Same shape (a square)

They differ only in size, by a factor of 2 in each linear dimension.

Step 6 — Notice the position changed too.

The original square sat in the region $[1, 3] \times [1, 3]$. The dilated square sits in $[2, 6] \times [2, 6]$. The figure didn't just grow; it also moved away from the origin, because every point's distance from the origin doubled.

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Answer: Dilating the unit square $(1, 1)$, $(3, 1)$, $(3, 3)$, $(1, 3)$ by $k = 2$ from the origin gives a square at $(2, 2)$, $(6, 2)$, $(6, 6)$, $(2, 6)$ — with side length 4 and area 16 ($= 4 \times \text{original area}$).

Lengths scale as $k$, areas as $k^{2}$, volumes as $k^{3}$.

Try It

• Adjust the scale factor $k$.
• Try $k = 0.5$ (reduction) — the square shrinks toward the origin.
• Try $k = -1$ — the square is rotated 180° about the origin.
• The HUD shows length, area, and the four vertex coordinates.