Draw the axes in the range −5≤x≤4 and −6≤y≤7 using a scale of 2cm to represent 1 unit on both axes. Draw and label triangle ABC with coordinates (1,1), (1,3) and (2,3) respectively.

The transformation is given by $H \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$.

For point $A(1,1)$:

$$A_6 = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} (1)(1) + (-2)(1) \\ (0)(1) + (1)(1) \end{pmatrix} = \begin{pmatrix} 1 - 2 \\ 0 + 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}$$

So, $A_6(-1,1)$.

For point $B(1,3)$:

$$B_6 = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} (1)(1) + (-2)(3) \\ (0)(1) + (1)(3) \end{pmatrix} = \begin{pmatrix} 1 - 6 \\ 0 + 3 \end{pmatrix} = \begin{pmatrix} -5 \\ 3 \end{pmatrix}$$

So, $B_6(-5,3)$.

For point $C(2,3)$:

$$C_6 = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} (1)(2) + (-2)(3) \\ (0)(2) + (1)(3) \end{pmatrix} = \begin{pmatrix} 2 - 6 \\ 0 + 3 \end{pmatrix} = \begin{pmatrix} -4 \\ 3 \end{pmatrix}$$

So, $C_6(-4,3)$.

The coordinates of triangle $A_6B_6C_6$ are $A_6(-1,1)$, $B_6(-5,3)$, and $C_6(-4,3)$.

(i) To draw and label triangle $A_6B_6C_6$, plot these points on the coordinate graph and connect them.

The matrix $H = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$ is a shear matrix.
A general shear matrix is of the form $\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ for a shear parallel to the x-axis, or $\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}$ for a shear parallel to the y-axis.
In this case, the matrix is $\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$, which corresponds to a shear parallel to the x-axis.
The line $y=0$ (the x-axis) is invariant under this transformation, as for any point $(x,0)$, $\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ 0 \end{pmatrix} = \begin{pmatrix} x \\ 0 \end{pmatrix}$.
The shear factor is $k = -2$.

The transformation $H$ is a shear.
The shear factor is $-2$.
The invariant line is the x-axis (or $y=0$).

The final answer is $\boxed{\text{f) (i) } A_6(-1,1), B_6(-5,3), C_6(-4,3) \text{ (Plot these points to draw the triangle.) (ii) Shear, shear factor -2, invariant line } y=0}$.