1. The gradient of line AB is \(\frac{q - 3}{p - 8}\).
2. The perpendicular bisector has a gradient of -2, so the gradient of AB is \(\frac{1}{2}\).
3. Set \(\frac{q - 3}{p - 8} = \frac{1}{2}\) and solve for q:
\(q - 3 = \frac{1}{2}(p - 8)\)
\(2q - p = -2\)
4. The midpoint of AB is \(\left( \frac{8 + p}{2}, \frac{3 + q}{2} \right)\).
5. Substitute the midpoint into the equation of the perpendicular bisector:
\(\frac{3 + q}{2} = -2 \left( \frac{8 + p}{2} \right) + 4\)
\(3 + q = -2(8 + p) + 8\)
\(3 + q = -16 - 2p + 8\)
\(q = -11 - 2p\)
6. Solve the system of equations:
\(2q - p = -2\)
\(q = -11 - 2p\)
Substitute \(q = -11 - 2p\) into \(2q - p = -2\):
\(2(-11 - 2p) - p = -2\)
\(-22 - 4p - p = -2\)
\(-5p = 20\)
\(p = -4\)
Substitute \(p = -4\) into \(q = -11 - 2p\):
\(q = -11 - 2(-4)\)
\(q = -11 + 8\)
\(q = -3\)