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Let's work through the problem step by step:

### Given:
The vertices of quadrilateral ABCD are:
- A = (-8, 8)
- B = (8, 3)
- C = (7, -1)
- D = (-4, 1)

The gradient of line AB is given as \(\frac{-5}{16}\).

### Part (a) - Calculate the gradient of line DC:
The formula for the gradient (slope) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
For points D \((-4, 1)\) and C \((7, -1)\):
\[
m_{DC} = \frac{-1 - 1}{7 - (-4)} = \frac{-2}{7 + 4} = \frac{-2}{11}
\]
So, the gradient of line DC is \(\frac{-2}{11}\).

### Part (b) - State whether or not DC is parallel to AB:
Lines are parallel if their gradients are equal. The gradient of AB is \(\frac{-5}{16}\), and the gradient of DC is \(\frac{-2}{11}\).

Since \(\frac{-5}{16} \neq \frac{-2}{11}\), the lines AB and DC are **not parallel**.

### Part (c) - Find the equation of the line through B and D:
The equation of a straight line can be written in the form \(y = mx + c\), where \(m\) is the gradient and \(c\) is the y-intercept. To find the equation of the line through B \((8, 3)\) and D \((-4, 1)\), we first calculate the gradient \(m_{BD}\):
\[
m_{BD} = \frac{3 - 1}{8 - (-4)} = \frac{2}{8 + 4} = \frac{2}{12} = \frac{1}{6}
\]
Now, using the point-slope form of the equation \(y - y_1 = m(x - x_1)\), with point \(B(8, 3)\):
\[
y - 3 = \frac{1}{6}(x - 8)
\]
Multiplying both sides by 6 to eliminate the fraction:
\[
6(y - 3) = x - 8
\]
\[
6y - 18 = x - 8
\]
Bringing all terms to one side:
\[
x - 6y = -10
\]
Thus, the equation of the line through B and D is:
\[
x - 6y = -10
\]

### Part (d) - Find the coordinates of the intersection point T of lines AC and BD:
We are given the equation of line AC as \(3x + 5y = 16\), and from part (c), the equation of line BD is \(x - 6y = -10\).

To find the intersection point \(T\), we need to solve this system of equations:
1. \(3x + 5y = 16\)
2. \(x - 6y = -10\)

I will solve these equations now.

The coordinates of the intersection point \( T \) of lines AC and BD are \( (2, 2) \).

### Summary of Solutions:
- **(a)** The gradient of the line DC is \(\frac{-2}{11}\).
- **(b)** The lines DC and AB are not parallel since their gradients are different.
- **(c)** The equation of the line through B and D is \( x - 6y = -10 \).
- **(d)** The coordinates of the intersection point \( T \) of lines AC and BD are \( (2, 2) \).

Would you like any further clarifications or details?

Here are 5 related questions to extend your understanding:
1. How would the equations change if the coordinates of point B were different?
2. How do you determine if two lines are perpendicular based on their gradients?
3. What is the significance of the y-intercept in a line equation?
4. Can two lines intersect at more than one point in a 2D plane?
5. How would you solve for the midpoint of a line segment like BD?

**Tip:** When solving simultaneous equations involving line intersections, substitution or elimination methods are the most efficient.

The vertices of quadrilateral ABCD are A (-8, 8), B (8, 3), C (7, -1), and D (-4, 1). Given that the gradient of AB is -5/16, calculate the gradient of DC, determine if DC is parallel to AB, find the equation of the line through B and D, and calculate the intersection point of lines AC and BD.

Solve the coordinate geometry problem involving quadrilateral ABCD with vertices A (-8, 8), B (8, 3), C (7, -1), and D (-4, 1). Calculate gradients, determine if lines are parallel, find line equations, and solve for the intersection of lines AC and BD.

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