Question

A(1, 2), B(2, 3) and C(4, 3) are the vertices of a $$\triangle \mathrm{ABC}$$.

Find:

1. the equation of altitude through B; 
2. the equation of altitude through C; 
3. the co-ordinates of the orthocentre of $$\triangle \mathrm{ABC}$$.

Answer 1

i. Altitude through $$\mathrm{B}$$ is perpendicular to $$\mathrm{AC}$$. 

Slope of $$\mathrm{AC}$$ joining $$\mathrm{A}(1, 2)$$ and $$\mathrm{C}(4, 3)$$: 

$$m_{\mathrm{AC}} = \frac{3 - 2}{4 - 1} = \frac{1}{3}$$ 

Slope of altitude through $$\mathrm{B}$$ is $$-\frac{1}{m_{\mathrm{AC}}} = -3$$. 

Passing through $$\mathrm{B}(2, 3)$$: 

$$y - 3 = -3(x - 2)$$

$$y - 3 = -3x + 6$$

$$3x + y - 9 = 0 \quad ...(1)$$

ii. Altitude through $$\mathrm{C}$$ is perpendicular to $$\mathrm{AB}$$.

Slope of $$\mathrm{AB}$$ joining $$\mathrm{A}(1, 2)$$ and $$\mathrm{B}(2, 3)$$:

$$m_{\mathrm{AB}} = \frac{3 - 2}{2 - 1}$$

$$= \frac{1}{1}$$

$$= 1$$

Slope of altitude through $$\mathrm{C}$$ is $$-\frac{1}{m_{\mathrm{AB}}} = -1$$. 

Passing through $$\mathrm{C}(4, 3)$$:

$$y - 3 = -1(x - 4)$$

$$y - 3 = -x + 4$$

$$x + y - 7 = 0 \quad ...(2)$$

iii. The orthocentre is the point of intersection of the altitudes.

Subtract (2) from (1):

$$(3x + y - 9) - (x + y - 7) = 0$$

$$2x - 2 = 0$$

$$\implies x = 1$$

Substitute $$x = 1$$ into (2):

$$1 + y - 7 = 0$$

$$\implies y = 6$$

Hence, the coordinates of the orthocentre are $$(1, 6)$$.