Plot the quadrilateral ABCD with vertices A(-6, -1), B(-6, -4), C(3, -7), D(3, 2). Then, draw the image A'B'C'D' under an enlargement with centre (0, -1) and scale factor 1/3.

23. Transformations of a Quadrilateral

First, plot the original quadrilateral ABCD.
Vertices: A(-6, -1), B(-6, -4), C(3, -7), D(3, 2).
Plot these points on the grid and connect them to form quadrilateral ABCD.

a) Draw the image A'B'C'D' under enlargement centre (0, -1) scale factor $\frac{1}{3}$.

To find the image coordinates $P'(x', y')$ of a point $P(x, y)$ under enlargement with centre $E(a, b)$ and scale factor $k$, use the formula:

$$P'(x', y') = (a + k(x-a), b + k(y-b))$$

Given centre $E(0, -1)$ and scale factor $k = \frac{1}{3}$.

For A(-6, -1):

$$A' = (0 + \frac{1}{3}(-6-0), -1 + \frac{1}{3}(-1 - (-1)))$$ $$A' = (0 + \frac{1}{3}(-6), -1 + \frac{1}{3}(0))$$ $$A' = (-2, -1)$$

For B(-6, -4):

$$B' = (0 + \frac{1}{3}(-6-0), -1 + \frac{1}{3}(-4 - (-1)))$$ $$B' = (0 + \frac{1}{3}(-6), -1 + \frac{1}{3}(-3))$$ $$B' = (-2, -1 - 1)$$ $$B' = (-2, -2)$$

For C(3, -7):

$$C' = (0 + \frac{1}{3}(3-0), -1 + \frac{1}{3}(-7 - (-1)))$$ $$C' = (0 + \frac{1}{3}(3), -1 + \frac{1}{3}(-6))$$ $$C' = (1, -1 - 2)$$ $$C' = (1, -3)$$

For D(3, 2):

$$D' = (0 + \frac{1}{3}(3-0), -1 + \frac{1}{3}(2 - (-1)))$$ $$D' = (0 + \frac{1}{3}(3), -1 + \frac{1}{3}(3))$$ $$D' = (1, -1 + 1)$$ $$D' = (1, 0)$$

The coordinates of the image A'B'C'D' are:

$$A'(-2, -1), B'(-2, -2), C'(1, -3), D'(1, 0)$$

Plot these points and connect them to form quadrilateral A'B'C'D'.

b) Draw A''B''C''D'' the image of A'B'C'D' under rotation of +ve 90° about (1, 0).

To find the image coordinates $P''(x'', y'')$ of a point $P'(x', y')$ under a +ve 90° (counter-clockwise) rotation about a centre $R(a, b)$, use the formula:

$$P''(x'', y'') = (a - (y'-b), b + (x'-a))$$

Given centre $R(1, 0)$.

For A'(-2, -1):

$$A'' = (1 - (-1-0), 0 + (-2-1))$$ $$A'' = (1 - (-1), 0 + (-3))$$ $$A'' = (2, -3)$$

For B'(-2, -2):

$$B'' = (1 - (-2-0), 0 + (-2-1))$$ $$B'' = (1 - (-2), 0 + (-3))$$ $$B'' = (3, -3)$$

For C'(1, -3):

$$C'' = (1 - (-3-0), 0 + (1-1))$$ $$C'' = (1 - (-3), 0 + 0)$$ $$C'' = (4, 0)$$

For D'(1, 0):

$$D'' = (1 - (0-0), 0 + (1-1))$$ $$D'' = (1 - 0, 0 + 0)$$ $$D'' = (1, 0)$$

The coordinates of the image A''B''C''D'' are:

$$A''(2, -3), B''(3, -3), C''(4, 0), D''(1, 0)$$

Plot these points and connect them to form quadrilateral A''B''C''D''.

c) Draw A'''B'''C'''D''' the image of A''B''C''D'' under a reflection in the line $y-x = 0$ (i.e., $y=x$).

To find the image coordinates $P'''(x''', y''')$ of a point $P''(x'', y'')$ under reflection in the line $y=x$, swap the coordinates:

$$P'''(x''', y''') = (y'', x'')$$

For A''(2, -3):

$$A''' = (-3, 2)$$

For B''(3, -3):

$$B''' = (-3, 3)$$

For C''(4, 0):

$$C''' = (0, 4)$$

For D''(1, 0):

$$D''' = (0, 1)$$

The coordinates of the image A'''B'''C'''D''' are:

$$A'''(-3, 2), B'''(-3, 3), C'''(0, 4), D'''(0, 1)$$

Plot these points and connect them to form quadrilateral A'''B'''C'''D'''.

d) Draw A$^{IV}$B$^{IV}$C$^{IV}$D$^{IV}$ the image of A'''B'''C'''D''' under translation $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and write the co-ordinate of the final image.

To find the image coordinates $P^{IV}(x^{IV}, y^{IV})$ of a point $P'''(x''', y''')$ under translation by vector $\begin{pmatrix} a \\ b \end{pmatrix}$, add the translation components to the coordinates:

$$P^{IV}(x^{IV}, y^{IV}) = (x''' + a, y''' + b)$$

Given translation vector $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$.

For A'''(-3, 2):

$$A^{IV} = (-3+2, 2+3)$$ $$A^{IV} = (-1, 5)$$

For B'''(-3, 3):

$$B^{IV} = (-3+2, 3+3)$$ $$B^{IV} = (-1, 6)$$

For C'''(0, 4):

$$C^{IV} = (0+2, 4+3)$$ $$C^{IV} = (2, 7)$$

For D'''(0, 1):

$$D^{IV} = (0+2, 1+3)$$ $$D^{IV} = (2, 4)$$

The coordinates of the final image A$^{IV}$B$^{IV}$C$^{IV}$D$^{IV}$ are:

$$\boxed{A^{IV}(-1, 5), B^{IV}(-1, 6), C^{IV}(2, 7), D^{IV}(2, 4)}$$

Plot these points and connect them to form quadrilateral A$^{IV}$B$^{IV}$C$^{IV}$D$^{IV}$.