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

(i) the equation of altitude through B

(ii) the equation of altitude through C

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

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

The altitude through B is perpendicular to the side AC.

Let the slope of altitude be m₁,

⇒ m_AC × m₂ = -1

⇒ $\dfrac{1}{3}$ × m₁ = -1

⇒ m₁ = -3

By point slope formula,

Equation of altitude B,

⇒ y - 3 = -3(x - 2)

⇒ y - 3 = -3x + 6

⇒ 3x + y - 9 = 0 …(1)

Hence, the equation of the altitude through B is 3x + y - 9 = 0.

(ii) Slope of AB = $\dfrac{y$2 - y1}{x2 - x1} = \dfrac{3 - 2}{2 - 1} = \dfrac{1}{1} = 1

The altitude through C is perpendicular to the side AB.

Let the slope of altitude be m₂,

⇒ m_AB × m₂ = -1

⇒ 1 × m₂ = -1

⇒ m₂ = -1

By point slope formula,

Equation of altitude B,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = -1(x - 4)

⇒ y - 3 = -x + 4

⇒ x + y - 7 = 0 …(2)

Hence, the equation of the altitude through C is x + y - 7 = 0.

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

⇒ (3x + y - 9) - (x + y - 7) = 0 - 0

⇒ 3x + y - 9 - x - y + 7 = 0 - 0

⇒ 3x - x + y - y - 9 + 7 = 0

⇒ 2x - 2 = 0

⇒ 2x = 2

⇒ x = 1

Substitute x = 1 into Equation (2):

⇒ 1 + y - 7 = 0

⇒ y - 6 = 0

⇒ y = 6.

Hence, the co-ordinates of the orthocentre of ΔABC are (1, 6).