The vertices of triangle PQR are P(3; 1), Q(4; 1) and R(4:3). Answer the whole of this question on a sheet of graph paper.

For the x-coordinates:
$3 \to -2$ (change of $-5$)
$4 \to -1$ (change of $-5$)
$4 \to -1$ (change of $-5$)
For the y-coordinates:
$1 \to -3$ (change of $-4$)
$1 \to -3$ (change of $-4$)
$3 \to -1$ (change of $-4$)

The transformation is a translation by the vector $\begin{pmatrix} -5 \\ -4 \end{pmatrix}$.

Part (c):
$\triangle P_2Q_2R_2$ is the image of $\triangle PQR$ under a reflection in the line $y=x$.
The rule for reflection in the line $y=x$ is $(x,y) \to (y,x)$.

The vertices of $\triangle P_2Q_2R_2$ are $P_2(1,3)$, $Q_2(1,4)$, and $R_2(3,4)$.
To draw $\triangle P_2Q_2R_2$, plot these points and connect them.

Part (d):
$\triangle PQR$ is enlarged with centre $(0,1)$ and scale factor $-\frac{1}{2}$ onto $\triangle P_3Q_3R_3$.
The formula for enlargement with centre $(a,b)$ and scale factor $k$ is $x' = a + k(x-a)$ and $y' = b + k(y-b)$. Here, $(a,b) = (0,1)$ and $k = -\frac{1}{2}$.

For $P(3,1)$:

For $Q(4,1)$:

For $R(4,3)$:

The vertices of $\triangle P_3Q_3R_3$ are $P_3\left(-\frac{3}{2}, 1\right)$, $Q_3(-2, 1)$, and $R_3(-2, 0)$.
To draw $\triangle P_3Q_3R_3$, plot these points and connect them.

Part (e):
A stretch represented by the matrix $\begin{pmatrix} 1 & 0 \\ 0 & -2 \end{pmatrix}$ maps $\triangle PQR$ onto $\triangle P_4Q_4R_4$.
The transformation rule is $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ -2y \end{pmatrix}$.

For $P(3,1)$:

For $Q(4,1)$:

For $R(4,3)$:

The vertices of $\triangle P_4Q_4R_4$ are $P_4(3, -2)$, $Q_4(4, -2)$, and $R_4(4, -6)$.
To draw $\triangle P_4Q_4R_4$, plot these points and connect them.