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Let's go through each part of this question one by one:

### (i) Find the equation of line \( AD \).

To find the equation of line \( AD \), we need to determine the slope of line \( AD \) and use the point-slope form of a line equation. We know:
- The coordinates of point \( A \) are \( (2, 6) \).
- We need to find the coordinates of point \( D \) next, but for now, we'll assume it lies on the line.

We are also told that \( AB \perp AD \), and angle \( BAD = 90^\circ \), meaning \( AD \) is vertical. A vertical line has an undefined slope and takes the form \( x = c \).

Since point \( A \) is \( (2, 6) \), the line \( AD \) must be the vertical line passing through \( x = 2 \). Therefore, the equation of the line \( AD \) is:

\[
x = 2
\]

---

### (ii) Find, by calculation, the coordinates of \( D \).

Since \( AB \parallel DC \) and \( AB \) is vertical, \( DC \) must also be vertical. From the trapezium’s properties, \( D \) must have the same \( x \)-coordinate as \( C \) because \( DC \) is vertical.

The coordinates of \( C \) are \( (8, 3) \), so the \( x \)-coordinate of \( D \) is also 8.

To find the \( y \)-coordinate of \( D \), note that \( AB \) is perpendicular to \( AD \), meaning that the vertical height between \( A \) and \( D \) is the same as the height between \( B \) and \( C \). The height difference between points \( B \) and \( C \) is:

\[
y_C - y_B = 3 - (-3) = 6
\]

Since \( D \) lies directly above point \( A \), the difference between \( A \)'s \( y \)-coordinate and \( D \)'s \( y \)-coordinate must also be 6. Therefore, the \( y \)-coordinate of \( D \) is:

\[
y_A + 6 = 6 + 6 = 12
\]

Thus, the coordinates of point \( D \) are \( (8, 12) \).

---

### (iii) Find the length of \( BE \).

We are told that point \( E \) is such that \( ABCE \) forms a parallelogram. For \( ABCE \) to be a parallelogram, \( E \) must be the translation of point \( C \) relative to point \( B \). The vector from \( B \) to \( C \) is:

\[
\vec{BC} = (8 - 5, 3 - (-3)) = (3, 6)
\]

Therefore, the coordinates of \( E \) are the result of applying the same vector \( \vec{BC} \) to point \( A \):

\[
E = A + \vec{BC} = (2 + 3, 6 + 6) = (5, 12)
\]

Now, to find the length of \( BE \), we use the distance formula between \( B(5, -3) \) and \( E(5, 12) \):

\[
BE = \sqrt{(5 - 5)^2 + (12 - (-3))^2} = \sqrt{0 + (12 + 3)^2} = \sqrt{15^2} = 15
\]

So, the length of \( BE \) is \( 15 \) units.

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Would you like a deeper explanation for any part? Here are some related questions to further your understanding:

1. How do you derive the equation of a line when it’s not vertical or horizontal?
2. What are the properties of a parallelogram, and how do they apply to coordinate geometry?
3. Why does the slope of a vertical line remain undefined, and what is the implication geometrically?
4. How can we find the equation of \( BC \), and does it play any role in solving for point \( E \)?
5. What is the significance of vectors in determining points like \( E \) in coordinate geometry?

**Tip**: Always sketch the diagram and mark the coordinates when dealing with geometry problems—this helps visualize the situation better!

The diagram shows a trapezium ABCD in which AB is parallel to DC and angle BAD is 90°. The coordinates of A, B, and C are (2, 6), (5, -3), and (8, 3) respectively.
(i) Find the equation of AD.
(ii) Find, by calculation, the coordinates of D.
The point E is such that ABCE is a parallelogram.
(iii) Find the length of BE.

Learn how to find the equation of a line, coordinates of a point, and the length between two points in a trapezium with given coordinates. This problem covers key concepts in coordinate geometry, properties of trapeziums and parallelograms, and is suitable for students in Grades 10-12.

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