**Transformation III, Long Questions (Question 2)**

**Question 2:**

**(a)**Transformation

**P**is a reflection in the line

*x*=

*m*.

Transformation

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

**R**is a clockwise rotation of 90^{o}about the centre (0, 4).(i) The point (6, 4) is the image of the point ( –2, 4) under the transformation

**P**.

State the value of

*m*.(ii) Find the coordinates of the image of point (2, 8) under the following combined transformations:

(a)

**T**,^{2}(b)

**TR**.(i)

*HEFG*is the image of*CDEF*under the combined transformation**WU**.Describe in full the transformation:

(a)

**U**(b)**W**(ii) It is given that

*CDEF*represents a region of area 60 m

^{2}.

Calculate the area, in m

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

*Solution:***(a)(i)**

$\begin{array}{l}\left(6,4\right)\to P\to \left(-2,4\right)\\ m=\frac{6+\left(-2\right)}{2}=2\end{array}$

**(a)(ii)**

**(b)**(2, 8) →

**R**→ (4, 2) →

**T**→ (8, 0)

(b)(i)(a)

(b)(i)(a)

**U**: An anticlockwise rotation of 90

^{o}about the centre

*A*(3, 3).

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

$\text{Scale factor}=\frac{HE}{CD}=\frac{4}{2}=2$

**W**: An enlargement of scale factor 2 with centre*B*(3, 5).

**(b)(ii)**

Area of

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

^{2}× area of*CDEF* = 4 × 60

= 240 m

^{2}Therefore,

Area of the shaded region

= Area of

*HEFG*– area of*CDEF*= 240 – 60

=

**180 m**^{2}