Given points: P, R(3; 5), S(-3; -7), T(-5; k). Q(0; 5) is a point on the y-axis. PT || RS. RS cuts the y-axis at D. PR cuts the x-axis at F. PT cuts the x-axis at E. angle PEF = θ. Write down the equa

From the diagram, $P$ is on the y-axis and has the same y-coordinate as $Q(0;5)$, so $P(0;5)$.
The coordinates of $R$ are given as $(3;5)$.

Since both points have the same y-coordinate ($y=5$), the line PR is a horizontal line.

The equation of PR is $\boxed{y=5}$.

3.2 Calculate the:

3.2.1 Gradient of RS

$R(3;5)$ and $S(-3;-7)$.

The gradient of RS is $\boxed{2}$.

3.2.2 Size of $\theta$

$\theta$ is the angle $\angle PEF$. This is the angle of inclination of the line PT with the positive x-axis.

It is given that $PT \parallel RS$. Therefore, their gradients are equal.
From 3.2.1, $m_{RS} = 2$.
So, $m_{PT} = 2$.

The gradient $m$ is equal to $\tan(\theta)$.

The size of $\theta$ is $\boxed{63.43^\circ}$.

3.2.3 Coordinates of D

$D$ is the y-intercept of the line RS.

We have the gradient $m_{RS} = 2$ and a point $R(3;5)$.
Using the point-slope form $y - y_1 = m(x - x_1)$:

Substitute $x=0$ into the equation:

The coordinates of D are $\boxed{(0;-1)}$.

3.3 If it is given that $TS = 2\sqrt{5}$, calculate the value of $k$.

$T(-5; k)$ and $S(-3; -7)$.

Given $TS = 2\sqrt{5}$.

Take the square root of both sides:

Case 1: $4 = 7+k$

Case 2: $-4 = 7+k$

From the diagram, point T is in the third quadrant, meaning its y-coordinate ($k$) must be negative. Both values are negative. However, looking at the diagram, T is below the x-axis and its y-coordinate appears to be greater than -7 (S's y-coordinate). If $k=-11$, T would be much lower than S. If $k=-3$, T is higher than S, which aligns with the visual representation.

The value of $k$ is $\boxed{-3}$.

3.4 Parallelogram TDNS, with N in the 4th quadrant, is drawn. Calculate the coordinates of N.

$T(-5;-3)$ (using $k=-3$ from 3.3)
$D(0;-1)$
$S(-3;-7)$
Let $N(x_N; y_N)$.

The midpoint of TS must be the same as the midpoint of DN.

Midpoint of TS:

Midpoint of DN:

The coordinates of N are $\boxed{(-8;-9)}$.
Self-correction: The question states N is in the 4th quadrant. A point in the 4th quadrant has a positive x-coordinate and a negative y-coordinate. My calculated N(-8;-9) is in the 3rd quadrant. This means the order of vertices for the parallelogram is important.

Let's re-evaluate using vector addition or the property that opposite sides are parallel and equal in length.
For parallelogram TDNS, $\vec{TD} = \vec{SN}$.
$D - T = N - S$
$(0 - (-5); -1 - (-3)) = (x_N - (-3); y_N - (-7))$
$(5; 2) = (x_N + 3; y_N + 7)$

Equating components:
$5 = x_N + 3 \implies x_N = 2$
$2 = y_N + 7 \implies y_N = -5$

So, $N(2;-5)$. This point is in the 4th quadrant (positive x, negative y), which matches the condition.

The coordinates of N are $\boxed{(2;-5)}$.

3.5 APRD is reflected about the y-axis to form APRD'. Calculate the size of $\angle RDR'$.

The problem statement mentions "trapezium PRST" and "parallelogram TDNS". It does not explicitly define point A. However, the question asks about APRD. Let's assume A is a point that forms a quadrilateral APRD. Given the context of reflections, we need the coordinates of R and D.
$R(3;5)$
$D(0;-1)$
$P(0;5)$

When a point $(x,y)$ is reflected about the y-axis, its image is $(-x,y)$.
$R(3;5) \implies R'(-3;5)$
$D(0;-1) \implies D'(0;-1)$ (D is on the y-axis, so it's invariant under y-axis reflection)

We need to find the angle formed by the lines DR and DR'.
The coordinates are $D(0;-1)$, $R(3;5)$, $R'(-3;5)$.

Calculate the gradient of DR ($m_{DR}$):

Calculate the gradient of DR' ($m_{DR'}$):

$\tan(\alpha) = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$.
Let $m_1 = m_{DR} = 2$ and $m_2 = m_{DR'} = -2$.

The size of $\angle RDR'$ is $\boxed{53.13^\circ}$.