Solutions to this question by accurate drawing will not be accepted. $ABCD$ is a trapezium where $AD$ is parallel to $BC$ and perpendicular to $AB$. $E$ is a point on $AD$ and $BE$ passes through the origin. Points $B$, $C$ and $D$ are $(3, -2)$, $(-2, -6)$ and $(-8, -2)$ respectively.

1. State that one pair of opposite sides is parallel

Since $E$ lies on $AD$ and it is given that $AD$ is parallel to $BC$, we have $ED \parallel BC$.

2. Calculate the gradient of the line $BE$

The line $BE$ passes through the origin $O(0, 0)$ and $B(3, -2)$.
Gradient of $BE = \frac{-2 - 0}{3 - 0} = -\frac{2}{3}$

3. Calculate the gradient of the line $CD$

Using $C(-2, -6)$ and $D(-8, -2)$:
Gradient of $CD = \frac{-2 - (-6)}{-8 - (-2)} = \frac{4}{-6} = -\frac{2}{3}$

4. Conclude that $BCDE$ is a parallelogram

Since the gradient of $BE$ equals the gradient of $CD$, $BE \parallel CD$.
Because both pairs of opposite sides are parallel ($ED \parallel BC$ and $BE \parallel CD$), $BCDE$ is a parallelogram.

1. Use the property of the parallelogram $BCDE$ to relate the coordinates of $E$ to known points

Since $BCDE$ is a parallelogram (from part i), the translation from $C$ to $B$ is the same as the translation from $D$ to $E$.
$\vec{CB} = \begin{pmatrix} 3 - (-2) \\ -2 - (-6) \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}$

2. Calculate the coordinates of $E$

$\vec{DE} = \vec{CB} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}$
Coordinates of $E = (-8 + 5, -2 + 4) = (-3, 2)$

1. Find the gradient of the line $AD$.

Since $AD$ is parallel to $BC$, their gradients are equal.
Gradient of $BC = \frac{-6 - (-2)}{-2 - 3} = \frac{-4}{-5} = \frac{4}{5}$
Gradient of $AD = \frac{4}{5}$

2. Find the equation of the line $AD$.

Using point $D(-8, -2)$ and gradient $\frac{4}{5}$:
$y - (-2) = \frac{4}{5}(x - (-8))$
$y = \frac{4}{5}x + \frac{32}{5} - 2$
$y = \frac{4}{5}x + \frac{22}{5}$

3. Find the equation of the line $AB$.

Since $AB$ is perpendicular to $AD$, the gradient of $AB$ is $-\frac{1}{\frac{4}{5}} = -\frac{5}{4}$.
Using point $B(3, -2)$:
$y - (-2) = -\frac{5}{4}(x - 3)$
$y = -\frac{5}{4}x + \frac{15}{4} - 2$
$y = -\frac{5}{4}x + \frac{7}{4}$

4. Solve the simultaneous equations to find the coordinates of $A$.

Equating the two expressions for $y$:
$\frac{4}{5}x + \frac{22}{5} = -\frac{5}{4}x + \frac{7}{4}$
Multiply the entire equation by $20$ to clear denominators:
$16x + 88 = -25x + 35$
$41x = -53$
$x = -\frac{53}{41} = -1\frac{12}{41}$

Substitute $x$ back into the equation for $AB$:
$y = -\frac{5}{4}\left(-\frac{53}{41}\right) + \frac{7}{4}$
$y = \frac{265}{164} + \frac{287}{164} = \frac{552}{164}$
$y = \frac{138}{41} = 3\frac{15}{41}$

Thus, the coordinates of $A$ are $\left(-1\frac{12}{41}, 3\frac{15}{41}\right)$ (shown).

1. Recall the coordinates of the vertices of the parallelogram

From the given information and earlier parts, the vertices of parallelogram $BCDE$ are $B(3, -2)$, $C(-2, -6)$, $D(-8, -2)$, and $E(-3, 2)$.

2. Apply the shoelace formula to find the area of the parallelogram

$\begin{aligned} \text{Area} &= \frac{1}{2} \left| \begin{matrix} 3 & -2 & -8 & -3 & 3 \\ -2 & -6 & -2 & 2 & -2 \end{matrix} \right| \\ &= \frac{1}{2} | [3(-6) + (-2)(-2) + (-8)(2) + (-3)(-2)] - [(-2)(-2) + (-6)(-8) + (-2)(-3) + (2)(3)] | \\ &= \frac{1}{2} | (-18 + 4 - 16 + 6) - (4 + 48 + 6 + 6) | \\ &= \frac{1}{2} | -24 - 64 | \\ &= \frac{1}{2} |-88| \\ &= 44 \end{aligned}$

1. Apply the change of base formula to convert the logarithm to base 3

$\log_y 27 = \frac{\log_3 27}{\log_3 y}$

2. Evaluate the numerator and substitute the given expression for the denominator

$\log_3 27 = 3$
$\log_y 27 = \frac{3}{p}$

1. Change the base of $\log_{27} x^3$ to base 3

$\log_{27} x^3 = \frac{\log_3 x^3}{\log_3 27} = \frac{3 \log_3 x}{3} = \log_3 x$

2. Change the base of $\frac{1}{\log_{5x} 3}$ to base 3

$\frac{1}{\log_{5x} 3} = \log_3 5x$

3. Substitute back into the equation and combine the logarithmic terms

$\log_3 x = \log_3 6 - \log_3 5x$
$\log_3 x + \log_3 5x = \log_3 6$
$\log_3 (x \cdot 5x) = \log_3 6$
$\log_3 (5x^2) = \log_3 6$

4. Solve for $x$ and reject the invalid root

$5x^2 = 6$
$x^2 = \frac{6}{5} = 1.2$
$x = \sqrt{1.2} \text{ or } x = -\sqrt{1.2}$
Since $x > 0$ for the logarithms to be defined, we reject $x = -\sqrt{1.2}$.
$x = \sqrt{1.2} \approx 1.0954 \approx 1.10$ (to 3 s.f.)