图中显示了一个三角形 PQR，其顶点为 P(9,10) 和 Q(1,-2)。点 R 位于 PQ 的垂直平分线上，且直线 QR 的方程为 $y = 8x - 10$。

1. Find the coordinates of the midpoint of PQ

Midpoint of $PQ = \left(\frac{9+1}{2}, \frac{10+(-2)}{2}\right) = (5, 4)$

2. Find the gradient of the line segment PQ

Gradient of $PQ = \frac{10 - (-2)}{9 - 1} = \frac{12}{8} = \frac{3}{2}$

3. Determine the gradient of the perpendicular bisector

Gradient of perpendicular bisector $= -\frac{1}{\frac{3}{2}} = -\frac{2}{3}$

4. Find the equation of the perpendicular bisector

$y - 4 = -\frac{2}{3}(x - 5)$
$3(y - 4) = -2(x - 5)$
$3y - 12 = -2x + 10$
$2x + 3y = 22$

1. Find the equation of the perpendicular bisector of $PQ$ (from part (a))

Midpoint of $PQ = \left(\frac{9+1}{2}, \frac{10+(-2)}{2}\right) = (5, 4)$
Gradient of $PQ = \frac{-2-10}{1-9} = \frac{-12}{-8} = \frac{3}{2}$
Gradient of perpendicular bisector = $-\frac{2}{3}$
Equation of perpendicular bisector: $y - 4 = -\frac{2}{3}(x - 5) \implies 2x + 3y = 22$

2. Solve simultaneously to find the coordinates of $R$

Substitute $y = 8x - 10$ into $2x + 3y = 22$:
$2x + 3(8x - 10) = 22$
$2x + 24x - 30 = 22$
$26x = 52 \implies x = 2$
Substitute $x = 2$ into $y = 8x - 10$:
$y = 8(2) - 10 = 6$
Coordinates of $R$ are $(2, 6)$.

1. Recall the coordinates of P, Q, and R from earlier parts

From the question and previous parts, the vertices are $P(9, 10)$, $Q(1, -2)$, and $R(2, 6)$.

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

$\text{Area} = \frac{1}{2} \left| \begin{matrix} 9 & 1 & 2 & 9 \\ 10 & -2 & 6 & 10 \end{matrix} \right|$

3. Evaluate the determinant to find the final area

$\text{Area} = \frac{1}{2} |(9(-2) + 1(6) + 2(10)) - (10(1) + (-2)(2) + 6(9))|$
$= \frac{1}{2} |(-18 + 6 + 20) - (10 - 4 + 54)|$
$= \frac{1}{2} |8 - 60|$
$= \frac{1}{2} |-52| = 26$