Given a rectangle ABCD, where point A is (-3, 1) and line AB is parallel to 3y − x = 4. a) Find the equation of line AB. b) Find the equation of line AD.

The given line is $3y - x = 4$.
Rearrange it into the slope-intercept form $y = mx + c$:

The gradient of this line is $m_1 = \frac{1}{3}$.

Since line AB is parallel to the given line, their gradients are equal.
So, the gradient of line AB is $m_{AB} = \frac{1}{3}$.

Line AB passes through point A $(-3, 1)$ and has a gradient of $\frac{1}{3}$.
Using the point-slope form $y - y_1 = m(x - x_1)$:

Multiply by 3 to clear the fraction:

Rearrange into the general form $Ax + By + C = 0$:

The equation of line AB is $\boxed{x - 3y + 6 = 0}$.

In a rectangle, adjacent sides are perpendicular. Therefore, line AD is perpendicular to line AB.
The gradient of line AB is $m_{AB} = \frac{1}{3}$.
The gradient of a line perpendicular to AB is the negative reciprocal of $m_{AB}$:

Line AD passes through point A $(-3, 1)$ and has a gradient of $-3$.
Using the point-slope form $y - y_1 = m(x - x_1)$:

Rearrange into the general form $Ax + By + C = 0$:

The equation of line AD is $\boxed{3x + y + 8 = 0}$.

In a rectangle, opposite sides are parallel. Therefore, line BC is parallel to line AD.
The gradient of line AD is $m_{AD} = -3$.
So, the gradient of line BC is $m_{BC} = -3$.

Line BC passes through point C $(2, 6)$ and has a gradient of $-3$.
Using the point-slope form $y - y_1 = m(x - x_1)$:

Rearrange into the form $Ax + By = C$:

To get $\frac{x}{a} + \frac{y}{b} = 1$, divide the entire equation by 12:

The equation of line BC is $\boxed{\frac{x}{4} + \frac{y}{12} = 1}$.

Equation of line CD:

In a rectangle, opposite sides are parallel. Therefore, line CD is parallel to line AB.
The gradient of line AB is $m_{AB} = \frac{1}{3}$.
So, the gradient of line CD is $m_{CD} = \frac{1}{3}$.

Line CD passes through point C $(2, 6)$ and has a gradient of $\frac{1}{3}$.
Using the point-slope form $y - y_1 = m(x - x_1)$:

Multiply by 3 to clear the fraction:

Rearrange into the form $Ax + By = C$:

To get $\frac{x}{a} + \frac{y}{b} = 1$, divide the entire equation by $-16$:

This can also be written as:

The equation of line CD is $\boxed{\frac{x}{-16} + \frac{y}{16/3} = 1}$.

Point B is the intersection of line AB and line BC.
From part (a), the equation of line AB is $x - 3y + 6 = 0 \implies x - 3y = -6 \quad (1)$.
From part (c), the equation of line BC is $3x + y = 12 \quad (2)$.

From equation (2), express $y$ in terms of $x$:

Substitute this expression for $y$ into equation (1):

Now substitute $x=3$ back into the expression for $y$:

The coordinates of B are $\boxed{(3, 3)}$.