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We are given the following key details from the diagram and problem:

1. \( A = (-1, 6) \) and \( E = (3, 4) \).
2. The gradient (slope) of \( AB \) is \( \frac{1}{3} \).
3. \( \angle AEB = 90^\circ \), meaning line segments \( AE \) and \( EB \) are perpendicular to each other.

### Step-by-Step Solution:

#### Step 1: Equation of Line \( AB \)
The gradient of \( AB \) is given as \( \frac{1}{3} \), so the equation of the line \( AB \) can be written using the point-slope form:
\[
y - y_1 = m(x - x_1)
\]
where \( m \) is the gradient, and \( (x_1, y_1) \) is a point on the line.

Substituting \( A(-1, 6) \) into the equation:
\[
y - 6 = \frac{1}{3}(x + 1)
\]
Simplifying:
\[
y - 6 = \frac{1}{3}x + \frac{1}{3}
\]
\[
y = \frac{1}{3}x + \frac{19}{3}
\]
Thus, the equation of line \( AB \) is:
\[
y = \frac{1}{3}x + \frac{19}{3}
\]

#### Step 2: Perpendicularity Condition (Gradient of Line \( EB \))
Since \( \angle AEB = 90^\circ \), the lines \( AE \) and \( EB \) are perpendicular. The gradient of line \( AE \) can be found using the formula for the gradient between two points:
\[
m_{AE} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 6}{3 - (-1)} = \frac{-2}{4} = -\frac{1}{2}
\]
For two perpendicular lines, the product of their slopes is \( -1 \):
\[
m_{EB} \times m_{AE} = -1
\]
\[
m_{EB} \times \left(-\frac{1}{2}\right) = -1
\]
\[
m_{EB} = 2
\]
Thus, the gradient of line \( EB \) is \( 2 \).

#### Step 3: Equation of Line \( EB \)
Using the gradient of \( EB \) and the point \( E(3, 4) \), we can write the equation of the line \( EB \) as:
\[
y - 4 = 2(x - 3)
\]
Simplifying:
\[
y - 4 = 2x - 6
\]
\[
y = 2x - 2
\]
Thus, the equation of line \( EB \) is:
\[
y = 2x - 2
\]

#### Step 4: Finding the Coordinates of \( B \)
To find the coordinates of \( B \), we solve the system of equations for \( AB \) and \( EB \):
1. \( y = \frac{1}{3}x + \frac{19}{3} \) (from \( AB \))
2. \( y = 2x - 2 \) (from \( EB \))

Equating the two expressions for \( y \):
\[
\frac{1}{3}x + \frac{19}{3} = 2x - 2
\]
Multiplying through by 3 to eliminate fractions:
\[
x + 19 = 6x - 6
\]
Simplifying:
\[
19 + 6 = 6x - x
\]
\[
25 = 5x
\]
\[
x = 5
\]

Substituting \( x = 5 \) into the equation of line \( EB \):
\[
y = 2(5) - 2 = 10 - 2 = 8
\]

Thus, the coordinates of point \( B \) are:
\[
B(5, 8)
\]

Would you like further details or have any other questions?

### Relative Questions:
1. How do you find the gradient between two points?
2. How can perpendicular lines be used to solve geometry problems?
3. What is the point-slope form of a line, and when is it used?
4. How can you verify if two lines are perpendicular?
5. How can you solve systems of linear equations algebraically?

### Tip:
Always remember that perpendicular lines have slopes whose product is \( -1 \), a useful property in geometry problems involving right angles.

The diagram shows a quadrilateral ABCD. The point E lies on AD such that angle AEB = 90°. The line EC is parallel to the x-axis and the line CD is parallel to the y-axis. The points A and E are (-1, 6) and (3, 4) respectively. Given that the gradient of AB is 1/3, find the coordinates of B.

This problem involves finding the coordinates of point B in quadrilateral ABCD, where angle AEB is 90°, and EC and CD are parallel to the axes. Using concepts of gradients and perpendicular lines, the problem is solved step-by-step.

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