The diagram shows a trapezium with vertices A(6, 8), B, C and D. The sides AB and DC are parallel and angle ABC is $90^\circ$. AB passes through the origin O and the length of AB is 15 units. The points C and D lie on the y-axis and x-axis respectively.

1. Find the length of OA

$OA = \sqrt{(6 - 0)^2 + (8 - 0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ units

2. Find the length of OB

Since $AB = 15$ and $O$ lies on $AB$ between $A$ and $B$, $OB = AB - OA = 15 - 10 = 5$ units.

3. Use similar triangles or vector scaling to find coordinates of B

Since $A, O, B$ are collinear and $O$ is between $A$ and $B$, the vector $\vec{OB}$ is in the opposite direction to $\vec{OA}$.
$\vec{OB} = -\frac{OB}{OA} \vec{OA} = -\frac{5}{10} \binom{6}{8} = -\frac{1}{2} \binom{6}{8} = \binom{-3}{-4}$.
Alternatively, by similar triangles, the coordinates of $B$ are $(-\frac{1}{2} \times 6, -\frac{1}{2} \times 8) = (-3, -4)$.

1. Find the gradient of the line AB.

Gradient of $AB = \frac{8 - (-4)}{6 - (-3)} = \frac{12}{9} = \frac{4}{3}$

2. Find the gradient of the line BC using the perpendicularity condition.

Since $\angle ABC = 90^\circ$, $BC$ is perpendicular to $AB$.
Gradient of $BC = -\frac{1}{\text{Gradient of } AB} = -\frac{3}{4}$

3. Find the equation of line BC and the coordinates of C.

Equation of $BC$ passing through $B(-3, -4)$:
$y - (-4) = -\frac{3}{4}(x - (-3))$
$y = -\frac{3}{4}x - \frac{9}{4} - 4$
$y = -\frac{3}{4}x - \frac{25}{4}$
Since $C$ lies on the $y$-axis, $x = 0$.
$y = -\frac{25}{4} = -6.25$
Coordinates of $C$ are $(0, -6.25)$.

4. Find the gradient of the line DC.

Since $DC$ is parallel to $AB$, they have the same gradient.
Gradient of $DC = \frac{4}{3}$

5. Find the equation of line DC and the coordinates of D.

Equation of $DC$ passing through $C(0, -\frac{25}{4})$:
$y = \frac{4}{3}x - \frac{25}{4}$
Since $D$ lies on the $x$-axis, $y = 0$.
$0 = \frac{4}{3}x - \frac{25}{4}$
$\frac{4}{3}x = \frac{25}{4}$
$x = \frac{25}{4} \times \frac{3}{4} = \frac{75}{16} = 4.6875$
Coordinates of $D$ are $(4.6875, 0)$.