**5.3b Determining the Image of an Object under Combination of (a) Two Enlargements or (b) an Enlargement and an Isometric Transformation**

**1.**

**Enlargement**is a transformation where all points of an object on a plane move from a fixed point at a constant ratio.

**2.**The fixed point is called the

**centre of enlargement**and constant ratio is called the

**scale factor**.

$\overline{)\text{Scale factor,}k=\frac{\text{Length of image side}}{\text{Length of object side}}\text{}}$

**3.**For enlargement, the object and the image are

**similar**.

**4.**Area of image = (Scale factor)

^{2}× Area of object

=

*k*^{2}× Area of object**Example:**

*E*,

*P*and

*T*are three transformations that are defined as follows:

*E*= Enlargement with centre

*V*(0, –1) and a scale factor of 2.

*P*= Reflection at the line

*y*= –1.

$T=Translation\text{}\left(\begin{array}{l}3\\ -3\end{array}\right)$

Based on the diagram above, determine the image of the shaded figure under the combined transformations

(a)

*E*^{2}(b)*ET*(c)*EP*

*Solution:*

**(a)**

Shaded figure → (E) figure

*III*→ (E) figure*D*.Hence, the image of the shaded figure under the combined transformation

*E*^{2 }=*EE*is the**figure**.*D*

**(b)**

Shaded figure → (T) figure

*II*→ (E) figure*A*.Hence, the image of the shaded figure under the combined transformation

*ET*is the**figure**.*A*

**(c)**

Shaded figure → (P) figure

*I*→ (E) figure*B*.Hence, the image of the shaded figure under the combined transformation

*EP*is the**figure**.*B*
P = should be y = -1

not x = -1

Dear Ricky,

thanks for pointing out our mistake, correction had been made accordingly.