Problem Statement
Three regional healthcare clinics are located at coordinates $A(2, 6), B(8, 8),$ and $C(6, 2)$ where distances are in kilometers.
**(a)** [3 marks] Find the equation of the perpendicular bisector of the line segment $AB$.
**(b)** [4 marks] Given that the perpendicular bisector of $BC$ has equation $y = \frac{1}{3}x + \frac{10}{3}$, find the coordinates of the circumcenter $V$ equidistant from $A, B,$ and $C$, and calculate the radius of the maximum empty circle centered at $V$.
Verified Solution & Marking Scheme
Part (a) [M1]
Midpoint of $AB$: $M_{AB} = \left(\frac{2+8}{2}, \frac{6+8}{2}\right) = (5, 7)$
Part (a) [M1]
Gradient of $AB$: $m_{AB} = \frac{8 - 6}{8 - 2} = \frac{2}{6} = \frac{1}{3} \implies m_\perp = -3$
Part (a) [A1]
Equation of bisector: $y - 7 = -3(x - 5) \implies y = -3x + 22$
Part (b) [M1]
Find intersection of $y = -3x + 22$ and $y = \frac{1}{3}x + \frac{10}{3}$:
$$-3x + 22 = \frac{1}{3}x + \frac{10}{3} \implies -9x + 66 = x + 10 \implies 10x = 56 \implies x = 5.6$$
Part (b) [A1]
$y = -3(5.6) + 22 = -16.8 + 22 = 5.2$. So $V(5.6, 5.2)$
Part (b) [M1]
Distance from $V(5.6, 5.2)$ to $A(2, 6)$:
$$R = \sqrt{(5.6 - 2)^2 + (5.2 - 6)^2} = \sqrt{3.6^2 + (-0.8)^2} = \sqrt{12.96 + 0.64} = \sqrt{13.6} \approx 3.69\text{ km}$$
Part (b) [A1]
Radius of largest empty circle: $R \approx 3.69\text{ km}$