**Question 9 (12 marks)**:

**(a)**Diagram 9.1 shows point

*P*(5, 1) on a Cartesian plane.

**Diagram 9.1**

Transformation

**T**is a translation $\left(\begin{array}{l}\text{}4\\ -3\end{array}\right)$

Transformation

**S**is an enlargement about the centre (–5, 2) with a scale factor 2.

State the coordinates of the image of point

*P*under the following transformations:

**(i) T**,

^{2}**(ii)**

**TS.**

**(b)**Diagram 9.2 shows geometrical shapes

*KLMNP*,

*KSRQP*and

*KTUVW*drawn on a Cartesian plane.

Diagram 9.2

(i)

Diagram 9.2

(i)

*KTUVW*is the image of

*KLMNP*under the combined transformation

**YZ**.

Describe, in full, the transformation:

**(a)**

**Z**,

**(b)**

**Y**.

**(ii)**It is given that

*KSRQP*represents a region of area 30 m

^{2}.

Calculate the area, in m

^{2}, of the shaded region.

*Solution:*

**(a)**

(i)

(i)

**TT**= P(–3, 3) →

**T**→ P’(1, 0) ) →

_{1}**T**→ P’’(5, –3)

_{2}**(ii) TS =**P (–3, 3) →

**S**→ P’(–1, 4) →

**T**→ P’’(3, 1)

**(b)(i)(a)**

**Z**: Reflection in the line

*x*= 0.

**(b)(i)(b)**

**Y**: Enlargement with the centre at (0, 0) and a scale factor of 2.

**(b)(ii)**

Area of

*KTUVW*= (Scale factor)

^{2}× Area of object

= 2

^{2}× 30

= 120 m

^{2}Hence,

Area of shaded region

= Area

*KTUVW*– Area

*KSRQP*

= 120 – 30

=

**90 m**

^{2}