**Further Practice:**

**Transformation III, Long Questions (Question 1)**

**Question 1:**

**(a)**Transformation

**T**is a translation $\left(\begin{array}{l}-4\\ \text{}2\end{array}\right)$ and transformation

**P**is an anticlockwise rotation of 90

^{o}about the centre (1, 0).

State the coordinates of the image of point (5, 1) under each of the following transformation:

(i) Translation

**T**,(ii) Rotation

**P**,(iii) Combined transformation

**T**.^{2}**(b)**Diagram below shows three quadrilaterals,

*ABCD*,

*EFGH*and

*JKLM*, drawn on a Cartesian plane.

(i)

*JKLM*is the image of*ABCD*under the combined transformation**VW**.Describe in full the transformation:

(a)

**W**(b)**V**(ii) It is given that quadrilateral

*ABCD*represents a region of area 18 m

^{2}.

Calculate the area, in m

^{2}, of the region represented by the shaded region.

*Solution:***(a)**

**(b)**

**(i)(a)**

**W**: A reflection in the line

*x*= –2

**(i)(b)**

**V**: An enlargement of scale factor 3 with centre (0, 4).

**(b)(ii)**

Area of

*EFGH*= area of*ABCD*= 18 m^{2}Area of

*JKLM*= (Scale factor)^{2}× Area of object = 3

^{2}× area of*EFGH* = 3

^{2}× 18 = 162 m

^{2}Therefore,

Area of the shaded region

= Area of

*JKLM*– area of*EFGH*= 162 – 18

=

**144 m**^{2}

**Question 2:**(a) Diagram below shows point

*A*and straight line

*y*+

*x*= 5 drawn on a Cartesian plane.

Transformation

**is a translation $\left(\begin{array}{l}\text{}5\\ -2\end{array}\right)$**

*T*Transformation

**is a reflection at the line**

*R**y*+

*x*= 5.

State the coordinates of the image of point

*A*under each of the following transformations:

**(i)**Transformation

**,**

*T***(ii)**Combined transformation

**.**

*TR***(b)**Diagram below shows pentagons

*JKLMN*,

*PQRST*and

*PUVWX*, drawn on a Cartesian plane.

(i)

*PUVWX*is the image of

*JKLMN*under the combined transformation

**CB**.

Describe in full the transformation:

(a)

**B**(b)

**C**

(ii) It is given that pentagon

*JKLMN*represents a region of area 80 m

^{2}.

Calculate the area, in m

^{2}, of the region represented by the shaded region.

$\begin{array}{l}\text{Scalefactor}=\frac{PU}{PQ}=\frac{6}{4}=\frac{3}{2}\\ \text{C:Anenlargementofscalefactor}\frac{3}{2}\text{withcentre}P\left(2,0\right)\end{array}$

(b)(ii)

Area of

= 80 m

$\begin{array}{l}\text{Areaof}PUVWX\\ ={\left(\frac{3}{2}\right)}^{2}\times \text{areaof}PQRST\\ =\frac{9}{4}\times 80\\ =180{\text{m}}^{2}\\ \\ \therefore \text{Areaoftheshadedregion}\\ \text{=areaof}PUVWX-\text{areaof}PQRST\\ \text{}=180-80\\ \text{}=100{\text{m}}^{2}\end{array}$

*Solution*:**(a)****(i)**(3, 4) → T → (8, 2)**(ii)**(3, 4) →**R**→ (1, 2) →**T**→ (6, 0)**(b)****(b)(i)(a)****B**: A clockwise rotation of 90^{o}about the centre (0, 2).**(b)(i)(b)**$\begin{array}{l}\text{Scalefactor}=\frac{PU}{PQ}=\frac{6}{4}=\frac{3}{2}\\ \text{C:Anenlargementofscalefactor}\frac{3}{2}\text{withcentre}P\left(2,0\right)\end{array}$

(b)(ii)

Area of

*PQRST*= Area of*JKLMN*= 80 m

^{2}$\begin{array}{l}\text{Areaof}PUVWX\\ ={\left(\frac{3}{2}\right)}^{2}\times \text{areaof}PQRST\\ =\frac{9}{4}\times 80\\ =180{\text{m}}^{2}\\ \\ \therefore \text{Areaoftheshadedregion}\\ \text{=areaof}PUVWX-\text{areaof}PQRST\\ \text{}=180-80\\ \text{}=100{\text{m}}^{2}\end{array}$