In the diagram, $AB$ is parallel to $DC$ and the coordinates of $A$, $C$ and $D$ are $(-4, 1)$, $(3, -1)$ and $(-1, -4)$ respectively. The gradient of line $OB$ is $\frac{7}{4}$. Find

1. Find the gradient of the line $DC$

Gradient of $DC = \frac{-1 - (-4)}{3 - (-1)} = \frac{3}{4}$

2. Form an equation for the line $AB$ using the fact that it is parallel to $DC$

Since $AB$ is parallel to $DC$, the gradient of $AB$ is also $\frac{3}{4}$.
Let the coordinates of $B$ be $(x, y)$.
Gradient of $AB = \frac{y - 1}{x - (-4)} = \frac{3}{4}$
$\frac{y - 1}{x + 4} = \frac{3}{4}$

3. Use the given gradient of $OB$ to form a second equation and solve for $x$ and $y$

Gradient of $OB = \frac{y - 0}{x - 0} = \frac{7}{4}$
$y = \frac{7}{4}x$

Substitute $y = \frac{7}{4}x$ into the equation for $AB$:
$\frac{\frac{7}{4}x - 1}{x + 4} = \frac{3}{4}$
$\frac{7}{4}x - 1 = \frac{3}{4}(x + 4)$
$\frac{7}{4}x - 1 = \frac{3}{4}x + 3$
$x = 4$

Substitute $x = 4$ back to find $y$:
$y = \frac{7}{4}(4) = 7$

The coordinates of $B$ are $(4, 7)$.

1. Retrieve the coordinates of $B$ from part (a) and set up the shoelace formula for the area of quadrilateral $ABCD$

From part (a), $B$ is $(4, 7)$.
Taking the vertices in counter-clockwise order $A(-4, 1)$, $D(-1, -4)$, $C(3, -1)$, $B(4, 7)$:
$\text{Area} = \frac{1}{2} \left| \begin{matrix} -4 & -1 & 3 & 4 & -4 \\ 1 & -4 & -1 & 7 & 1 \end{matrix} \right|$

2. Evaluate the determinant to find the area

$\text{Area} = \frac{1}{2} | [(-4)(-4) + (-1)(-1) + (3)(7) + (4)(1)] - [(1)(-1) + (-4)(3) + (-1)(4) + (7)(-4)] |$
$= \frac{1}{2} | (16 + 1 + 21 + 4) - (-1 - 12 - 4 - 28) |$
$= \frac{1}{2} | 42 - (-45) |$
$= \frac{1}{2} (87) = 43.5$

1. Set up an equation using the condition that $P$ is equidistant from $C$ and $D$.

Let the coordinates of $P$ be $(2.5, y)$.
Since $PC = PD$, $PC^2 = PD^2$.
$(2.5 - 3)^2 + (y - (-1))^2 = (2.5 - (-1))^2 + (y - (-4))^2$

2. Expand and solve for $y$.

$(-0.5)^2 + (y + 1)^2 = (3.5)^2 + (y + 4)^2$
$0.25 + y^2 + 2y + 1 = 12.25 + y^2 + 8y + 16$
$2y + 1.25 = 8y + 28.25$
$6y = -27$
$y = -4.5$

3. State the coordinates of $P$.

The coordinates of $P$ are $(2.5, -4.5)$.

1. Recall or calculate the coordinates of $B$ from part (a)

Gradient of $DC = \frac{-1 - (-4)}{3 - (-1)} = \frac{3}{4}$.
Since $AB \parallel DC$, gradient of $AB = \frac{3}{4}$.
Equation of $AB$: $y - 1 = \frac{3}{4}(x + 4) \implies y = \frac{3}{4}x + 4$.
Equation of $OB$: $y = \frac{7}{4}x$.
Solving simultaneously: $\frac{7}{4}x = \frac{3}{4}x + 4 \implies x = 4$.
$y = \frac{7}{4}(4) = 7$.
So, $B$ is $(4, 7)$.

2. Use the property that diagonals of a parallelogram bisect each other to find $C'$

For $ABC'D$ to be a parallelogram, the midpoint of diagonal $AC'$ must be the same as the midpoint of diagonal $BD$.
Midpoint of $BD = \left(\frac{4 + (-1)}{2}, \frac{7 + (-4)}{2}\right) = \left(\frac{3}{2}, \frac{3}{2}\right)$.
Let $C'$ be $(x, y)$.
Midpoint of $AC' = \left(\frac{-4 + x}{2}, \frac{1 + y}{2}\right)$.
Equating the midpoints:
$\frac{-4 + x}{2} = \frac{3}{2} \implies -4 + x = 3 \implies x = 7$
$\frac{1 + y}{2} = \frac{3}{2} \implies 1 + y = 3 \implies y = 2$
Therefore, $C'$ is $(7, 2)$.