Enlargement of shapes from center point A using scale factors.

To solve this problem, we need to enlarge each shape from the centre of enlargement A using the given scale factors. This is a transformation in geometry known as enlargement (or dilation).

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🔷 What is Enlargement?

- Centre of enlargement (A): The fixed point from which all points of the shape are scaled.
- Scale factor: How much bigger or smaller the new shape will be compared to the original.
- If scale factor > 1: enlargement
- If scale factor < 1: reduction
- If negative: image is on the opposite side of the centre

We use the following steps for each shape:

1. Identify the coordinates of the vertices of the shape.
2. Find the vector from the centre A to each vertex.
3. Multiply that vector by the scale factor.
4. Add the result back to the coordinates of A to get the new vertex.
5. Connect the new vertices to form the enlarged shape.

Let’s go through each one step-by-step. Since I can’t see the exact grid positions, I’ll assume typical coordinate placements based on standard grid layouts and describe how to do it. You can apply this method to your actual image.

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🔹 Shape 1: Triangle (Top Left)

- Scale Factor: 2
- Centre A: Below the triangle
- Steps:
- Let’s say the triangle has vertices at (2,6), (2,4), (6,4) and A is at (7,3).
- Vector from A to each vertex:
- (2–7, 6–3) = (-5,3) → ×2 = (-10,6) → New point: (7–10, 3+6) = (-3,9)
- (2–7,4–3)=(-5,1) → ×2=(-10,2) → (7–10,3+2)=(-3,5)
- (6–7,4–3)=(-1,1) → ×2=(-2,2) → (7–2,3+2)=(5,5)
- Draw triangle with vertices at (-3,9), (-3,5), (5,5)

✔ Result: Larger triangle, twice the size, same orientation, centered around A.

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🔹 Shape 2: Triangle (Top Middle)

- Scale Factor: 4
- Centre A: To the left of the triangle
- Steps:
- Suppose triangle vertices: (8,6), (10,6), (9,8), A at (5,6)
- Vector from A to (8,6): (3,0) → ×4 = (12,0) → New: (5+12,6+0) = (17,6)
- (10,6): (5,0) → ×4 = (20,0) → (25,6)
- (9,8): (4,2) → ×4 = (16,8) → (21,14)
- New triangle: (17,6), (25,6), (21,14)

✔ Much larger triangle, far from original, same direction.

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🔹 Shape 3: Square (Top Right)

- Scale Factor: 2
- Centre A: Below square
- Suppose square corners: (6,8), (8,8), (8,6), (6,6), A at (7,4)
- Vectors:
- (6–7,8–4)=(-1,4) → ×2=(-2,8) → (7–2,4+8)=(5,12)
- (8–7,8–4)=(1,4) → ×2=(2,8) → (9,12)
- (8–7,6–4)=(1,2) → ×2=(2,4) → (9,8)
- (6–7,6–4)=(-1,2) → ×2=(-2,4) → (5,8)
- New square: (5,12), (9,12), (9,8), (5,8)

✔ Double size, above A.

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🔹 Shape 4: Diamond (Rhombus) (Top Right)

- Scale Factor: 3
- Centre A: To the right of diamond
- Suppose diamond: (8,6), (9,7), (8,8), (7,7), A at (11,6)
- Vectors:
- (8–11,6–6)=(-3,0) → ×3=(-9,0) → (11–9,6+0)=(2,6)
- (9–11,7–6)=(-2,1) → ×3=(-6,3) → (5,9)
- (8–11,8–6)=(-3,2) → ×3=(-9,6) → (2,12)
- (7–11,7–6)=(-4,1) → ×3=(-12,3) → (-1,9)
- New diamond: (2,6), (5,9), (2,12), (-1,9)

✔ Tripled size, farther from A.

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🔹 Shape 5: Hexagon (Bottom Left)

- Scale Factor: 3
- Centre A: Below hexagon
- Assume regular hexagon centered near (5,5), A at (5,1)
- Each vertex is 2 units away vertically/horizontally from center
- Use vectors from A to each vertex, multiply by 3, add back to A
- Example: if a vertex is at (5,7), vector is (0,6), ×3 = (0,18), new point: (5,1+18)=(5,19)
- Repeat for all 6 vertices

✔ Large hexagon, stretched from A.

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🔹 Shape 6: L-shape (Bottom Middle Left)

- Scale Factor: 2
- Centre A: To the right
- Suppose L-shape has corners: (5,5), (7,5), (7,7), (5,7), (5,9), (3,9), (3,7)
- A at (9,6)
- For each point:
- Vector from A: e.g., (5–9,5–6)=(-4,-1) → ×2=(-8,-2) → (9–8,6–2)=(1,4)
- Continue for all vertices
- All points move away from A, doubled distance

✔ Larger L-shape, symmetrically expanded.

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🔹 Shape 7: T-shaped (Bottom Middle Right)

- Scale Factor: 4
- Centre A: Bottom right corner
- Suppose shape has points like (6,6), (8,6), (8,8), (6,8), (6,10), (4,10), (4,8)
- A at (8,6)
- Vector from A to (6,6): (-2,0) → ×4 = (-8,0) → (0,6)
- (8,6): (0,0) → stays at (8,6)
- (8,8): (0,2) → ×4 = (0,8) → (8,14)
- (6,8): (-2,2) → ×4 = (-8,8) → (0,14)
- etc.
- New shape is much larger, extending left and up

✔ Big T-shape, stretched from A.

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🔹 Shape 8: Parallelogram (Bottom Right)

- Scale Factor: 1½ = 3/2
- Centre A: At bottom right
- Suppose parallelogram: (5,4), (8,4), (9,6), (6,6), A at (9,4)
- Vectors:
- (5–9,4–4)=(-4,0) → ×1.5 = (-6,0) → (3,4)
- (8–9,4–4)=(-1,0) → ×1.5 = (-1.5,0) → (7.5,4)
- (9–9,6–4)=(0,2) → ×1.5 = (0,3) → (9,7)
- (6–9,6–4)=(-3,2) → ×1.5 = (-4.5,3) → (4.5,7)
- New points: (3,4), (7.5,4), (9,7), (4.5,7)

✔ Slightly larger, same orientation.

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✔ Summary of Steps to Solve Any Enlargement:

1. Identify Centre A — mark its coordinates.
2. Find coordinates of each vertex of the shape.
3. For each vertex:
- Calculate vector from A to vertex: (x – Ax, y – Ay)
- Multiply vector by scale factor
- Add result to A’s coordinates:
New x = Ax + (x – Ax) × k
New y = Ay + (y – Ay) × k
4. Plot new points, join them in order.
5. Check: The new shape should be similar, oriented the same way (unless scale factor negative), and distances from A scaled accordingly.

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📌 Final Tips:

- Use graph paper or digital grid.
- Always draw lines from A through each vertex to find where the new point lies.
- Scale factor 1½ means each point moves 1.5 times further from A.
- Negative scale factors flip the shape across A.

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Since you have the image, apply these steps manually on your grid. Use rulers or dot counting to ensure accuracy.

Would you like me to generate a diagram or help you with specific coordinates?