Question

$$\mathrm{A}(1, 2), \mathrm{B}(3, -4)$$ and $$\mathrm{C}(5, -6)$$ are the vertices of $$\triangle \mathrm{ABC}$$. 

Find: 

1. the equation of the right bisector of BC; 
2. the equation of the right bisector of CA; 
3. the co-ordinates of the circumcentre of $$\triangle \mathrm{ABC}$$.

Answer 1

i. Mid-point of $$\mathrm{BC}$$: 

$$\mathrm{D} = \left(\frac{3 + 5}{2}, \frac{-4 + (-6)}{2}\right)$$

$$= (4, -5)$$ 

Slope of $$\mathrm{BC}$$:

$$m_{\mathrm{BC}} = \frac{-6 - (-4)}{5 - 3}$$ 

$$= \frac{-2}{2}$$

$$= -1$$ 

Slope of right bisector of $$\mathrm{BC}$$ is $$-\frac{1}{-1} = 1$$. 

Equation of right bisector of $$\mathrm{BC}$$: 

$$y - (-5) = 1(x - 4)$$

$$y + 5 = x - 4$$

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

ii. Mid-point of $$\mathrm{CA}$$:

$$\mathrm{E} = \left(\frac{5 + 1}{2}, \frac{-6 + 2}{2}\right)$$

$$= (3, -2)$$ 

Slope of $$\mathrm{CA}$$: 

$$m_{\mathrm{CA}} = \frac{2 - (-6)}{1 - 5}$$

$$= \frac{8}{-4}$$

$$= -2$$ 

Slope of right bisector of $$\mathrm{CA}$$ is $$-\frac{1}{-2} = \frac{1}{2}$$. 

Equation of right bisector of $$\mathrm{CA}$$: 

$$y - (-2) = \frac{1}{2}(x - 3)$$

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

$$2y + 4 = x - 3$$

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

iii. The circumcentre is the intersection point of the right bisectors.

Subtract (2) from (1): 

$$(x - y - 9) - (x - 2y - 7) = 0$$ 

$$y - 2 = 0$$

$$\implies y = 2$$

Substitute $$y = 2$$ into (1): 

$$x - 2 - 9 = 0$$

$$\implies x = 11$$

Thus, the circumcentre of $$\triangle \mathrm{ABC}$$ is $$(11, 2)$$.