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.