A(1, 2), B(3, –4) and C(5, –6) are the vertices of ΔABC. Find :

(i) the equation of the right bisector of BC

(ii) the equation of the right bisector of CA

(iii) the co-ordinates of the circumcentre of ΔABC

(i) Slope of BC = $\dfrac{y$2 - y1}{x2 - x1} = \dfrac{-6 - (-4)}{5 - 3} = \dfrac{-2}{2} = -1

The right bisector is perpendicular to BC.

Let the slope of right bisector be m₁,

⇒ m_AB × m₁ = -1

⇒ -1 × m₁ = -1

⇒ m₁ = 1

The right bisector (perpendicular bisector) passes through the midpoint of BC

$x, y = \Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big) \\[1em] = \Big(\dfrac{3 + 5}{2}, \dfrac{-4 + (-6)}{2}\Big) \\[1em] = \Big(\dfrac{8}{2}, \dfrac{-10}{2}\Big) = (4, -5).

By point slope formula,

Equation of right bisector of BC,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-5) = 1(x - 4)

⇒ y + 5 = x - 4

⇒ x - y - 9 = 0….(1)

Hence, equation of right bisector of BC is x - y - 9 = 0.

(ii) Slope of CA = $\dfrac{y$2 - y1}{x2 - x1} = \dfrac{2 - (-6)}{1 - 5} = \dfrac{8}{-4} = -2

The right bisector is perpendicular to CA.

Let the slope of right bisector of CA be m₂,

⇒ m_CA × m₂ = -1

⇒ -2 × m₂ = -1

⇒ m₂ = $\dfrac{1}{2}$

Midpoint of CA

$x, y = \Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big) \\[1em] = \Big(\dfrac{5 + 1}{2}, \dfrac{-6 + 2 }{2}\Big) \\[1em] = \Big(\dfrac{6}{2}, \dfrac{-4}{2}\Big) = (3, -2).

By point slope formula,

Equation of right bisector of CA,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-2) = $\dfrac{1}{2}$ (x - 3)

⇒ 2(y + 2) = (x - 3)

⇒ 2y + 4 = (x - 3)

⇒ x - 2y - 7 = 0 …(2)

Hence, equation of right bisector of AC is x - 2y - 7 = 0.

(iii) Subtract Equation (2) from Equation (1):

⇒ (x - y) - (x - 2y) = 9 - 7

⇒ x - x - y + 2y = 2

⇒ y = 2.

Substitute y = 2 into Equation (1):

⇒ x - 2 = 9

⇒ x = 11.

Hence, coordinates of the circumcenter are (11, 2).