Enlargements

• The magnitude of $k$ determines the size of the image relative to the object. If $|k| > 1$, the image is larger than the object; if $0 < |k| < 1$, the image is smaller (often called a reduction).

• A positive scale factor ($k > 0$) means the image is on the same side of the centre of enlargement as the object and maintains the same orientation.

• A negative scale factor ($k < 0$) means the image is on the opposite side of the centre of enlargement and appears rotated by $180^\circ$ (inverted).

• The distance from the CoE to any point on the image is exactly $k$ times the distance from the CoE to the corresponding point on the object.

The Coordinate Method

• Step 1: Identify the coordinates of the Centre of Enlargement $(x_c, y_c)$ and a vertex of the object $(x_o, y_o)$.
• Step 2: Calculate the horizontal and vertical displacements from the CoE to the vertex: $dx = x_o - x_c$ and $dy = y_o - y_c$.
• Step 3: Multiply these displacements by the scale factor $k$: $dx' = k \cdot dx$ and $dy' = k \cdot dy$.
• Step 4: Add these new displacements back to the CoE coordinates to find the image vertex: $(x_c + dx', y_c + dy')$.

The Ray Method

• Draw straight lines (rays) starting from the CoE and passing through each vertex of the object. The corresponding vertices of the enlarged image must lie on these same rays.

• This method is highly effective for verifying the accuracy of a transformation or for finding the CoE when both the object and image are given.

• It is vital to distinguish between length scale factor and area scale factor. If the lengths are enlarged by $k$, the area is enlarged by $k^2$.

• Measuring from the Origin: Students often mistakenly measure distances from $(0,0)$ instead of the specified Centre of Enlargement. Always mark the CoE clearly first.

• Miscalculating k: Remember the formula is $\text{Scale Factor} = \frac{\text{Image Length}}{\text{Object Length}}$. Reversing this will result in the reciprocal of the correct answer.

• Area Confusion: Doubling the side lengths ($k=2$) quadruples the area ($k^2=4$). Many students incorrectly assume the area increases by the same factor as the lengths.