The diagram below, which is not drawn to scale, shows a triangle $PQR$ with vertices $P(-3, 9)$, $Q(3, 7)$ and $R(2, -1)$. The point $S$ which lies on $PR$ is the foot of the perpendicular line from $Q$.

1. Find the gradient of the line $PR$.

Gradient of $PR = \frac{9 - (-1)}{-3 - 2} = \frac{10}{-5} = -2$

2. Find the gradient of the perpendicular line $QS$.

Since $QS \perp PR$, gradient of $QS = -\frac{1}{-2} = \frac{1}{2}$

3. Find the equation of the line $QS$ passing through $Q(3, 7)$.

$y - 7 = \frac{1}{2}(x - 3)$
$y = \frac{1}{2}x - \frac{3}{2} + 7$
$y = \frac{1}{2}x + \frac{11}{2}$

1. Find the gradient of the line $PR$

Gradient of $PR = \frac{-1 - 9}{2 - (-3)} = \frac{-10}{5} = -2$

2. Find the equation of the line parallel to $PR$ passing through $Q$

Since the line is parallel to $PR$, its gradient is $-2$.
Using the point $Q(3, 7)$:
$y - 7 = -2(x - 3)$
$y - 7 = -2x + 6$
$y = -2x + 13$

1. Apply the shoelace formula to find the area of the triangle

$\text{Area} = \frac{1}{2} \left| \begin{matrix} -3 & 3 & 2 & -3 \\ 9 & 7 & -1 & 9 \end{matrix} \right|$

2. Evaluate the determinant to find the final area

$\text{Area} = \frac{1}{2} | (-3)(7) + (3)(-1) + (2)(9) - (9)(3) - (7)(2) - (-1)(-3) |$
$= \frac{1}{2} | -21 - 3 + 18 - 27 - 14 - 3 |$
$= \frac{1}{2} | -50 |$
$= 25$